-4f^2+48f+68=0

Solution for -4f^2+48f+68=0 equation:

-4f^2+48f+68=0

a = -4; b = 48; c = +68;
Δ = b2-4ac
Δ = 482-4·(-4)·68
Δ = 3392
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3392}=\sqrt{64*53}=\sqrt{64}*\sqrt{53}=8\sqrt{53}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(48)-8\sqrt{53}}{2*-4}=\frac{-48-8\sqrt{53}}{-8}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(48)+8\sqrt{53}}{2*-4}=\frac{-48+8\sqrt{53}}{-8}
$