10y2-4y-5=3y

To remove the radical on the left side of the equation, square both sides of the equation.

10y2-4y-52=(3y)2

Multiply the exponents in ((10y2-4y-5)12)2.

Apply the power rule and multiply exponents, (am)n=amn.

(10y2-4y-5)12⋅2=(3y)2

Cancel the common factor of 2.

Cancel the common factor.

(10y2-4y-5)12⋅2=(3y)2

Rewrite the expression.

(10y2-4y-5)1=(3y)2

(10y2-4y-5)1=(3y)2

(10y2-4y-5)1=(3y)2

Simplify.

10y2-4y-5=(3y)2

Apply the product rule to 3y.

10y2-4y-5=32y2

Raise 3 to the power of 2.

10y2-4y-5=9y2

10y2-4y-5=9y2

Move all terms containing y to the left side of the equation.

Subtract 9y2 from both sides of the equation.

10y2-4y-5-9y2=0

Subtract 9y2 from 10y2.

y2-4y-5=0

y2-4y-5=0

Factor y2-4y-5 using the AC method.

Consider the form x2+bx+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.

-5,1

Write the factored form using these integers.

(y-5)(y+1)=0

(y-5)(y+1)=0

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

y-5=0

y+1=0

Set the first factor equal to 0 and solve.

Set the first factor equal to 0.

y-5=0

Add 5 to both sides of the equation.

y=5

y=5

Set the next factor equal to 0 and solve.

Set the next factor equal to 0.

y+1=0

Subtract 1 from both sides of the equation.

y=-1

y=-1

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

y=5,-1

y=5,-1

Exclude the solutions that do not make 10y2-4y-5=3y true.

y=5

Solve for y square root of 10y^2-4y-5=3y