(y+12)(y-3)=(y+5)+24

Solution for (y+12)(y-3)=(y+5)+24 equation:

(y+12)(y-3)=(y+5)+24

We move all terms to the left:

(y+12)(y-3)-((y+5)+24)=0

We multiply parentheses ..

(+y^2-3y+12y-36)-((y+5)+24)=0


We calculate terms in parentheses: -((y+5)+24), so:

(y+5)+24

We get rid of parentheses

y+5+24

We add all the numbers together, and all the variables

y+29

Back to the equation:

-(y+29)


We get rid of parentheses

y^2-3y+12y-y-36-29=0

We add all the numbers together, and all the variables

y^2+8y-65=0

a = 1; b = 8; c = -65;
Δ = b2-4ac
Δ = 82-4·1·(-65)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-18}{2*1}=\frac{-26}{2}
=-13
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+18}{2*1}=\frac{10}{2}
=5
$