2x^2=24x+82

Solution for 2x^2=24x+82 equation:

2x^2=24x+82

We move all terms to the left:

2x^2-(24x+82)=0

We get rid of parentheses

2x^2-24x-82=0

a = 2; b = -24; c = -82;
Δ = b2-4ac
Δ = -242-4·2·(-82)
Δ = 1232
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:

$\sqrt{\Delta}=\sqrt{1232}=\sqrt{16*77}=\sqrt{16}*\sqrt{77}=4\sqrt{77}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-4\sqrt{77}}{2*2}=\frac{24-4\sqrt{77}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+4\sqrt{77}}{2*2}=\frac{24+4\sqrt{77}}{4}
$