(y+5)(3y-2)=0

Solution for (y+5)(3y-2)=0 equation:

(y+5)(3y-2)=0

We multiply parentheses ..

(+3y^2-2y+15y-10)=0

We get rid of parentheses

3y^2-2y+15y-10=0

We add all the numbers together, and all the variables

3y^2+13y-10=0

a = 3; b = 13; c = -10;
Δ = b2-4ac
Δ = 132-4·3·(-10)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-17}{2*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*3}=\frac{4}{6}
=2/3
$