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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

2F^2+2F-8=0

a = 2; b = 2; c = -8;
Δ = b2-4ac
Δ = 22-4·2·(-8)
Δ = 68
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{68}=\sqrt{4*17}=\sqrt{4}*\sqrt{17}=2\sqrt{17}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{17}}{2*2}=\frac{-2-2\sqrt{17}}{4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{17}}{2*2}=\frac{-2+2\sqrt{17}}{4}
$