2x^2-18x-20=48

Solution for 2x^2-18x-20=48 equation:

2x^2-18x-20=48

We move all terms to the left:

2x^2-18x-20-(48)=0

We add all the numbers together, and all the variables

2x^2-18x-68=0

a = 2; b = -18; c = -68;
Δ = b2-4ac
Δ = -182-4·2·(-68)
Δ = 868
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:

$\sqrt{\Delta}=\sqrt{868}=\sqrt{4*217}=\sqrt{4}*\sqrt{217}=2\sqrt{217}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{217}}{2*2}=\frac{18-2\sqrt{217}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{217}}{2*2}=\frac{18+2\sqrt{217}}{4}
$