F(x)=4x^2+32x-48

Solution for F(x)=4x^2+32x-48 equation:

(F)=4F^2+32F-48

We move all terms to the left:

(F)-(4F^2+32F-48)=0

We get rid of parentheses

-4F^2+F-32F+48=0

We add all the numbers together, and all the variables

-4F^2-31F+48=0

a = -4; b = -31; c = +48;
Δ = b2-4ac
Δ = -312-4·(-4)·48
Δ = 1729
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-31)-\sqrt{1729}}{2*-4}=\frac{31-\sqrt{1729}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1729}}{2*-4}=\frac{31+\sqrt{1729}}{-8}
$