(8f)^2=4f

Solution for (8f)^2=4f equation:

(8f)^2=4f

We move all terms to the left:

(8f)^2-(4f)=0

a = 8; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·8·0
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{-(-4)-4}{2*8}=\frac{0}{16}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*8}=\frac{8}{16}
=1/2
$