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

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

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

We multiply parentheses ..

(+5y^2-15y+4y-12)=0

We get rid of parentheses

5y^2-15y+4y-12=0

We add all the numbers together, and all the variables

5y^2-11y-12=0

a = 5; b = -11; c = -12;
Δ = b2-4ac
Δ = -112-4·5·(-12)
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*5}=\frac{-8}{10}
=-4/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+19}{2*5}=\frac{30}{10}
=3
$