Solve for q 5-4q=1q^2

$5 - 4 q = 1 q^{2}$

Multiply $q^{2}$ by $1$ .

$5 - 4 q = q^{2}$

Subtract $q^{2}$ from both sides of the equation.

$5 - 4 q - q^{2} = 0$

Factor the left side of the equation.

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Factor $- 1$ out of $5 - 4 q - q^{2}$ .

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Reorder the expression.

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Move $5$ .

$- 4 q - q^{2} + 5 = 0$

Reorder $- 4 q$ and $- q^{2}$ .

$- q^{2} - 4 q + 5 = 0$

Factor $- 1$ out of $- q^{2}$ .

$- (q^{2}) - 4 q + 5 = 0$

Factor $- 1$ out of $- 4 q$ .

$- (q^{2}) - (4 q) + 5 = 0$

Rewrite $5$ as $- 1 (- 5)$ .

$- (q^{2}) - (4 q) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (q^{2}) - (4 q)$ .

$- (q^{2} + 4 q) - 1 \cdot - 5 = 0$

Factor $- 1$ out of $- (q^{2} + 4 q) - 1 (- 5)$ .

$- (q^{2} + 4 q - 5) = 0$

Factor.

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Factor $q^{2} + 4 q - 5$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$- ((q - 1) (q + 5)) = 0$

Remove unnecessary parentheses.

$- (q - 1) (q + 5) = 0$

Multiply each term in $- (q - 1) (q + 5) = 0$ by $- 1$

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Multiply each term in $- (q - 1) (q + 5) = 0$ by $- 1$ .

$(- (q - 1) (q + 5)) \cdot - 1 = 0 \cdot - 1$

Simplify $(- (q - 1) (q + 5)) \cdot - 1$ .

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Simplify by multiplying through.

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Apply the distributive property.

$(- q - - 1) (q + 5) \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 1$ .

$(- q + 1) (q + 5) \cdot - 1 = 0 \cdot - 1$

Expand $(- q + 1) (q + 5)$ using the FOIL Method.

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Apply the distributive property.

$(- q (q + 5) + 1 (q + 5)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- q \cdot q - q \cdot 5 + 1 (q + 5)) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$(- q \cdot q - q \cdot 5 + 1 q + 1 \cdot 5) \cdot - 1 = 0 \cdot - 1$

Simplify and combine like terms.

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Simplify each term.

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Multiply $q$ by $q$ by adding the exponents.

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Move $q$ .

$(- (q \cdot q) - q \cdot 5 + 1 q + 1 \cdot 5) \cdot - 1 = 0 \cdot - 1$

Multiply $q$ by $q$ .

$(- q^{2} - q \cdot 5 + 1 q + 1 \cdot 5) \cdot - 1 = 0 \cdot - 1$

Multiply $5$ by $- 1$ .

$(- q^{2} - 5 q + 1 q + 1 \cdot 5) \cdot - 1 = 0 \cdot - 1$

Multiply $q$ by $1$ .

$(- q^{2} - 5 q + q + 1 \cdot 5) \cdot - 1 = 0 \cdot - 1$

Multiply $5$ by $1$ .

$(- q^{2} - 5 q + q + 5) \cdot - 1 = 0 \cdot - 1$

Add $- 5 q$ and $q$ .

$(- q^{2} - 4 q + 5) \cdot - 1 = 0 \cdot - 1$

Apply the distributive property.

$- q^{2} \cdot - 1 - 4 q \cdot - 1 + 5 \cdot - 1 = 0 \cdot - 1$

Simplify.

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Multiply $- q^{2} \cdot - 1$ .

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Multiply $- 1$ by $- 1$ .

$1 q^{2} - 4 q \cdot - 1 + 5 \cdot - 1 = 0 \cdot - 1$

Multiply $q^{2}$ by $1$ .

$q^{2} - 4 q \cdot - 1 + 5 \cdot - 1 = 0 \cdot - 1$

Multiply $- 1$ by $- 4$ .

$q^{2} + 4 q + 5 \cdot - 1 = 0 \cdot - 1$

Multiply $5$ by $- 1$ .

$q^{2} + 4 q - 5 = 0 \cdot - 1$

Multiply $0$ by $- 1$ .

$q^{2} + 4 q - 5 = 0$

Factor $q^{2} + 4 q - 5$ using the AC method.

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Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 5$ and whose sum is $4$ .

$- 1, 5$

Write the factored form using these integers.

$(q - 1) (q + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$q - 1 = 0$

$q + 5 = 0$

Set the first factor equal to $0$ and solve.

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Set the first factor equal to $0$ .

$q - 1 = 0$

Add $1$ to both sides of the equation.

$q = 1$

Set the next factor equal to $0$ and solve.

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Set the next factor equal to $0$ .

$q + 5 = 0$

Subtract $5$ from both sides of the equation.

$q = - 5$

The final solution is all the values that make $(q - 1) (q + 5) = 0$ true.

$q = 1, - 5$

Solve for q 5-4q=1q^2

Solving MATH problems