24x+2=2x^2+8

Solution for 24x+2=2x^2+8 equation:

24x+2=2x^2+8

We move all terms to the left:

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

We get rid of parentheses

-2x^2+24x-8+2=0

We add all the numbers together, and all the variables

-2x^2+24x-6=0

a = -2; b = 24; c = -6;
Δ = b2-4ac
Δ = 242-4·(-2)·(-6)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-4\sqrt{33}}{2*-2}=\frac{-24-4\sqrt{33}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+4\sqrt{33}}{2*-2}=\frac{-24+4\sqrt{33}}{-4}
$