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

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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+3x-1=0

a = 6; b = 3; c = -1;
Δ = b2-4ac
Δ = 32-4·6·(-1)
Δ = 33
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{33}}{2*6}=\frac{-3-\sqrt{33}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*6}=\frac{-3+\sqrt{33}}{12}
$