F(x)=2x^2-10x+10

Solution for F(x)=2x^2-10x+10 equation:

(F)=2F^2-10F+10

We move all terms to the left:

(F)-(2F^2-10F+10)=0

We get rid of parentheses

-2F^2+F+10F-10=0

We add all the numbers together, and all the variables

-2F^2+11F-10=0

a = -2; b = 11; c = -10;
Δ = b2-4ac
Δ = 112-4·(-2)·(-10)
Δ = 41
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{-(11)-\sqrt{41}}{2*-2}=\frac{-11-\sqrt{41}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+\sqrt{41}}{2*-2}=\frac{-11+\sqrt{41}}{-4}
$