2f2=2

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Solution for 2f2=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 $

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