2x^2-8x+18=48

Solution for 2x^2-8x+18=48 equation:

2x^2-8x+18=48

We move all terms to the left:

2x^2-8x+18-(48)=0

We add all the numbers together, and all the variables

2x^2-8x-30=0

a = 2; b = -8; c = -30;
Δ = b2-4ac
Δ = -82-4·2·(-30)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{19}}{2*2}=\frac{8-4\sqrt{19}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{19}}{2*2}=\frac{8+4\sqrt{19}}{4}
$