k(2k+5)=3

Simplify by multiplying through.

Apply the distributive property.

k(2k)+k⋅5=3

Reorder.

Rewrite using the commutative property of multiplication.

2k⋅k+k⋅5=3

Move 5 to the left of k.

2k⋅k+5⋅k=3

2k⋅k+5⋅k=3

2k⋅k+5⋅k=3

Multiply k by k by adding the exponents.

Move k.

2(k⋅k)+5⋅k=3

Multiply k by k.

2k2+5⋅k=3

2k2+5k=3

2k2+5k=3

Move 3 to the left side of the equation by subtracting it from both sides.

2k2+5k-3=0

For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=2⋅-3=-6 and whose sum is b=5.

Factor 5 out of 5k.

2k2+5(k)-3=0

Rewrite 5 as -1 plus 6

2k2+(-1+6)k-3=0

Apply the distributive property.

2k2-1k+6k-3=0

2k2-1k+6k-3=0

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

(2k2-1k)+6k-3=0

Factor out the greatest common factor (GCF) from each group.

k(2k-1)+3(2k-1)=0

k(2k-1)+3(2k-1)=0

Factor the polynomial by factoring out the greatest common factor, 2k-1.

(2k-1)(k+3)=0

(2k-1)(k+3)=0

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

2k-1=0

k+3=0

Set the first factor equal to 0.

2k-1=0

Add 1 to both sides of the equation.

2k=1

Divide each term by 2 and simplify.

Divide each term in 2k=1 by 2.

2k2=12

Cancel the common factor of 2.

Cancel the common factor.

2k2=12

Divide k by 1.

k=12

k=12

k=12

k=12

Set the next factor equal to 0.

k+3=0

Subtract 3 from both sides of the equation.

k=-3

k=-3

The final solution is all the values that make (2k-1)(k+3)=0 true.

k=12,-3

The result can be shown in multiple forms.

Exact Form:

k=12,-3

Decimal Form:

k=0.5,-3

Solve for k k(2k+5)=3