(y+5)(3y-4)=0

Solution for (y+5)(3y-4)=0 equation:

(y+5)(3y-4)=0

We multiply parentheses ..

(+3y^2-4y+15y-20)=0

We get rid of parentheses

3y^2-4y+15y-20=0

We add all the numbers together, and all the variables

3y^2+11y-20=0

a = 3; b = 11; c = -20;
Δ = b2-4ac
Δ = 112-4·3·(-20)
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-19}{2*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+19}{2*3}=\frac{8}{6}
=1+1/3
$