2x^2-14x+24x=x-32x-8

Solution for 2x^2-14x+24x=x-32x-8 equation:

2x^2-14x+24x=x-32x-8

We move all terms to the left:

2x^2-14x+24x-(x-32x-8)=0

We add all the numbers together, and all the variables

2x^2-14x+24x-(-31x-8)=0

We add all the numbers together, and all the variables

2x^2+10x-(-31x-8)=0

We get rid of parentheses

2x^2+10x+31x+8=0

We add all the numbers together, and all the variables

2x^2+41x+8=0

a = 2; b = 41; c = +8;
Δ = b2-4ac
Δ = 412-4·2·8
Δ = 1617
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:

$\sqrt{\Delta}=\sqrt{1617}=\sqrt{49*33}=\sqrt{49}*\sqrt{33}=7\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(41)-7\sqrt{33}}{2*2}=\frac{-41-7\sqrt{33}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(41)+7\sqrt{33}}{2*2}=\frac{-41+7\sqrt{33}}{4}
$