F(4x+1)(x-8)=8

Solution for F(4x+1)(x-8)=8 equation:

(4F+1)(F-8)=8

We move all terms to the left:

(4F+1)(F-8)-(8)=0

We multiply parentheses ..

(+4F^2-32F+F-8)-8=0

We get rid of parentheses

4F^2-32F+F-8-8=0

We add all the numbers together, and all the variables

4F^2-31F-16=0

a = 4; b = -31; c = -16;
Δ = b2-4ac
Δ = -312-4·4·(-16)
Δ = 1217
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{1217}}{2*4}=\frac{31-\sqrt{1217}}{8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-31)+\sqrt{1217}}{2*4}=\frac{31+\sqrt{1217}}{8}
$