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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2x^2-9x+6=0

a = -2; b = -9; c = +6;
Δ = b2-4ac
Δ = -92-4·(-2)·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{-(-9)-\sqrt{129}}{2*-2}=\frac{9-\sqrt{129}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{129}}{2*-2}=\frac{9+\sqrt{129}}{-4}
$