(2x^2+3x)=0

Solution for (2x^2+3x)=0 equation:

(2x^2+3x)=0

We get rid of parentheses

2x^2+3x=0

a = 2; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·2·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*2}=\frac{-6}{4}
=-1+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*2}=\frac{0}{4}
=0
$