(2x^2+3x-5)-(-2x^2-3x+5)=0

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

(2x^2+3x-5)-(-2x^2-3x+5)=0

We get rid of parentheses

2x^2+2x^2+3x+3x-5-5=0

We add all the numbers together, and all the variables

4x^2+6x-10=0

a = 4; b = 6; c = -10;
Δ = b2-4ac
Δ = 62-4·4·(-10)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-14}{2*4}=\frac{-20}{8}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+14}{2*4}=\frac{8}{8}
=1
$