F(b)=5(b)^2+6(b)-9

Solution for F(b)=5(b)^2+6(b)-9 equation:

(F)=5(F)^2+6(F)-9

We move all terms to the left:

(F)-(5(F)^2+6(F)-9)=0

We get rid of parentheses

-5F^2+F-6F+9=0

We add all the numbers together, and all the variables

-5F^2-5F+9=0

a = -5; b = -5; c = +9;
Δ = b2-4ac
Δ = -52-4·(-5)·9
Δ = 205
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{-(-5)-\sqrt{205}}{2*-5}=\frac{5-\sqrt{205}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{205}}{2*-5}=\frac{5+\sqrt{205}}{-10}
$