(2y-1)(y+4)=y2+7y

Solution for (2y-1)(y+4)=y2+7y equation:

(2y-1)(y+4)=y2+7y

We move all terms to the left:

(2y-1)(y+4)-(y2+7y)=0

We add all the numbers together, and all the variables

-(+y^2+7y)+(2y-1)(y+4)=0

We get rid of parentheses

-y^2-7y+(2y-1)(y+4)=0

We multiply parentheses ..

-y^2+(+2y^2+8y-1y-4)-7y=0

We add all the numbers together, and all the variables

-1y^2+(+2y^2+8y-1y-4)-7y=0

We get rid of parentheses

-1y^2+2y^2+8y-1y-7y-4=0

We add all the numbers together, and all the variables

y^2-4=0

a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2}
=-2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2}
=2
$