2x^2-8=41

Solution for 2x^2-8=41 equation:

2x^2-8=41

We move all terms to the left:

2x^2-8-(41)=0

We add all the numbers together, and all the variables

2x^2-49=0

a = 2; b = 0; c = -49;
Δ = b2-4ac
Δ = 02-4·2·(-49)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-14\sqrt{2}}{2*2}=\frac{0-14\sqrt{2}}{4}
=-\frac{14\sqrt{2}}{4}
=-\frac{7\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+14\sqrt{2}}{2*2}=\frac{0+14\sqrt{2}}{4}
=\frac{14\sqrt{2}}{4}
=\frac{7\sqrt{2}}{2}
$