(y-3)(y+8)=0

Solution for (y-3)(y+8)=0 equation:

(y-3)(y+8)=0

We multiply parentheses ..

(+y^2+8y-3y-24)=0

We get rid of parentheses

y^2+8y-3y-24=0

We add all the numbers together, and all the variables

y^2+5y-24=0

a = 1; b = 5; c = -24;
Δ = b2-4ac
Δ = 52-4·1·(-24)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-11}{2*1}=\frac{-16}{2}
=-8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+11}{2*1}=\frac{6}{2}
=3
$