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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -4; b = -8; c = +5;
Δ = b2-4ac
Δ = -82-4·(-4)·5
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12}{2*-4}=\frac{-4}{-8}
=1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12}{2*-4}=\frac{20}{-8}
=-2+1/2
$