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

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

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

We multiply parentheses ..

(+2y^2-6y+5y-15)=0

We get rid of parentheses

2y^2-6y+5y-15=0

We add all the numbers together, and all the variables

2y^2-1y-15=0

a = 2; b = -1; c = -15;
Δ = b2-4ac
Δ = -12-4·2·(-15)
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{-(-1)-11}{2*2}=\frac{-10}{4}
=-2+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+11}{2*2}=\frac{12}{4}
=3
$