32=y(y+4)

Solution for 32=y(y+4) equation:

32=y(y+4)

We move all terms to the left:

32-(y(y+4))=0


We calculate terms in parentheses: -(y(y+4)), so:

y(y+4)

We multiply parentheses

y^2+4y

Back to the equation:

-(y^2+4y)


We get rid of parentheses

-y^2-4y+32=0

We add all the numbers together, and all the variables

-1y^2-4y+32=0

a = -1; b = -4; c = +32;
Δ = b2-4ac
Δ = -42-4·(-1)·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-12}{2*-1}=\frac{-8}{-2}
=+4
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+12}{2*-1}=\frac{16}{-2}
=-8
$