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

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

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

We multiply parentheses ..

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

We get rid of parentheses

2y^2-6y+y-3=0

We add all the numbers together, and all the variables

2y^2-5y-3=0

a = 2; b = -5; c = -3;
Δ = b2-4ac
Δ = -52-4·2·(-3)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*2}=\frac{-2}{4}
=-1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*2}=\frac{12}{4}
=3
$