F(x)=4x^2-26x+40

Solution for F(x)=4x^2-26x+40 equation:

(F)=4F^2-26F+40

We move all terms to the left:

(F)-(4F^2-26F+40)=0

We get rid of parentheses

-4F^2+F+26F-40=0

We add all the numbers together, and all the variables

-4F^2+27F-40=0

a = -4; b = 27; c = -40;
Δ = b2-4ac
Δ = 272-4·(-4)·(-40)
Δ = 89
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{-(27)-\sqrt{89}}{2*-4}=\frac{-27-\sqrt{89}}{-8}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+\sqrt{89}}{2*-4}=\frac{-27+\sqrt{89}}{-8}
$