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

Solution for 3x^2+2x-4=-2x^2-x+2 equation:

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

5x^2+3x-6=0

a = 5; b = 3; c = -6;
Δ = b2-4ac
Δ = 32-4·5·(-6)
Δ = 129
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{129}}{2*5}=\frac{-3-\sqrt{129}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{129}}{2*5}=\frac{-3+\sqrt{129}}{10}
$