(y-3)(y+2)=4y

Solution for (y-3)(y+2)=4y equation:

(y-3)(y+2)=4y

We move all terms to the left:

(y-3)(y+2)-(4y)=0

We add all the numbers together, and all the variables

-4y+(y-3)(y+2)=0

We multiply parentheses ..

(+y^2+2y-3y-6)-4y=0

We get rid of parentheses

y^2+2y-3y-4y-6=0

We add all the numbers together, and all the variables

y^2-5y-6=0

a = 1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*1}=\frac{-2}{2}
=-1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*1}=\frac{12}{2}
=6
$