10=2x^2

Solution for 10=2x^2 equation:

10=2x^2

We move all terms to the left:

10-(2x^2)=0

a = -2; b = 0; c = +10;
Δ = b2-4ac
Δ = 02-4·(-2)·10
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{5}}{2*-2}=\frac{0-4\sqrt{5}}{-4}
=-\frac{4\sqrt{5}}{-4}
=-\frac{\sqrt{5}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{5}}{2*-2}=\frac{0+4\sqrt{5}}{-4}
=\frac{4\sqrt{5}}{-4}
=\frac{\sqrt{5}}{-1}
$