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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

x^2-8x+4=0

a = 1; b = -8; c = +4;
Δ = b2-4ac
Δ = -82-4·1·4
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{3}}{2*1}=\frac{8-4\sqrt{3}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{3}}{2*1}=\frac{8+4\sqrt{3}}{2}
$