F(3)=-x^2-5x+36

Solution for F(3)=-x^2-5x+36 equation:

(3)=-F^2-5F+36

We move all terms to the left:

(3)-(-F^2-5F+36)=0

We get rid of parentheses

F^2+5F-36+3=0

We add all the numbers together, and all the variables

F^2+5F-33=0

a = 1; b = 5; c = -33;
Δ = b2-4ac
Δ = 52-4·1·(-33)
Δ = 157
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{157}}{2*1}=\frac{-5-\sqrt{157}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{157}}{2*1}=\frac{-5+\sqrt{157}}{2}
$