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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2x^2-1x-13=0

a = 2; b = -1; c = -13;
Δ = b2-4ac
Δ = -12-4·2·(-13)
Δ = 105
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{-(-1)-\sqrt{105}}{2*2}=\frac{1-\sqrt{105}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{105}}{2*2}=\frac{1+\sqrt{105}}{4}
$