5(2y+1)(3y-5)=0

Solution for 5(2y+1)(3y-5)=0 equation:

5(2y+1)(3y-5)=0

We multiply parentheses ..

5(+6y^2-10y+3y-5)=0

We multiply parentheses

30y^2-50y+15y-25=0

We add all the numbers together, and all the variables

30y^2-35y-25=0

a = 30; b = -35; c = -25;
Δ = b2-4ac
Δ = -352-4·30·(-25)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-65}{2*30}=\frac{-30}{60}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+65}{2*30}=\frac{100}{60}
=1+2/3
$