0.1=2x^2

Solution for 0.1=2x^2 equation:

0.1=2x^2

We move all terms to the left:

0.1-(2x^2)=0

a = -2; b = 0; c = +0.1;
Δ = b2-4ac
Δ = 02-4·(-2)·0.1
Δ = 0.8
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{0.8}}{2*-2}=\frac{0-\sqrt{0.8}}{-4}
=-\frac{\sqrt{}}{-4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{0.8}}{2*-2}=\frac{0+\sqrt{0.8}}{-4}
=\frac{\sqrt{}}{-4}
$