F(2)=5x^2+10x

Solution for F(2)=5x^2+10x equation:

(2)=5F^2+10F

We move all terms to the left:

(2)-(5F^2+10F)=0

We get rid of parentheses

-5F^2-10F+2=0

a = -5; b = -10; c = +2;
Δ = b2-4ac
Δ = -102-4·(-5)·2
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*-5}=\frac{10-2\sqrt{35}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*-5}=\frac{10+2\sqrt{35}}{-10}
$