F(x)=4x^2-80x+395

Solution for F(x)=4x^2-80x+395 equation:

(F)=4F^2-80F+395

We move all terms to the left:

(F)-(4F^2-80F+395)=0

We get rid of parentheses

-4F^2+F+80F-395=0

We add all the numbers together, and all the variables

-4F^2+81F-395=0

a = -4; b = 81; c = -395;
Δ = b2-4ac
Δ = 812-4·(-4)·(-395)
Δ = 241
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{-(81)-\sqrt{241}}{2*-4}=\frac{-81-\sqrt{241}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(81)+\sqrt{241}}{2*-4}=\frac{-81+\sqrt{241}}{-8}
$