3x^2-2x+4=5x^2+6x-8

Solution for 3x^2-2x+4=5x^2+6x-8 equation:

3x^2-2x+4=5x^2+6x-8

We move all terms to the left:

3x^2-2x+4-(5x^2+6x-8)=0

We get rid of parentheses

3x^2-5x^2-2x-6x+8+4=0

We add all the numbers together, and all the variables

-2x^2-8x+12=0

a = -2; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·(-2)·12
Δ = 160
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:

$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{10}}{2*-2}=\frac{8-4\sqrt{10}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{10}}{2*-2}=\frac{8+4\sqrt{10}}{-4}
$