2x^2-5=20

Solution for 2x^2-5=20 equation:

2x^2-5=20

We move all terms to the left:

2x^2-5-(20)=0

We add all the numbers together, and all the variables

2x^2-25=0

a = 2; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·2·(-25)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*2}=\frac{0-10\sqrt{2}}{4}
=-\frac{10\sqrt{2}}{4}
=-\frac{5\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*2}=\frac{0+10\sqrt{2}}{4}
=\frac{10\sqrt{2}}{4}
=\frac{5\sqrt{2}}{2}
$