F(x)=4x^2-32x+68

Solution for F(x)=4x^2-32x+68 equation:

(F)=4F^2-32F+68

We move all terms to the left:

(F)-(4F^2-32F+68)=0

We get rid of parentheses

-4F^2+F+32F-68=0

We add all the numbers together, and all the variables

-4F^2+33F-68=0

a = -4; b = 33; c = -68;
Δ = b2-4ac
Δ = 332-4·(-4)·(-68)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-1}{2*-4}=\frac{-34}{-8}
=4+1/4
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+1}{2*-4}=\frac{-32}{-8}
=+4
$