2f^2=2

Solution for 2f^2=2 equation:

2f^2=2

We move all terms to the left:

2f^2-(2)=0

a = 2; b = 0; c = -2;
Δ = b2-4ac
Δ = 02-4·2·(-2)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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