(2x^2-3x-2)-(4x^2+5x+3)=0

Solution for (2x^2-3x-2)-(4x^2+5x+3)=0 equation:

(2x^2-3x-2)-(4x^2+5x+3)=0

We get rid of parentheses

2x^2-4x^2-3x-5x-2-3=0

We add all the numbers together, and all the variables

-2x^2-8x-5=0

a = -2; b = -8; c = -5;
Δ = b2-4ac
Δ = -82-4·(-2)·(-5)
Δ = 24
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{6}}{2*-2}=\frac{8-2\sqrt{6}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{6}}{2*-2}=\frac{8+2\sqrt{6}}{-4}
$