F(x)=-x^2+4x+5

Solution for F(x)=-x^2+4x+5 equation:

(F)=-F^2+4F+5

We move all terms to the left:

(F)-(-F^2+4F+5)=0

We get rid of parentheses

F^2-4F+F-5=0

We add all the numbers together, and all the variables

F^2-3F-5=0

a = 1; b = -3; c = -5;
Δ = b2-4ac
Δ = -32-4·1·(-5)
Δ = 29
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{-(-3)-\sqrt{29}}{2*1}=\frac{3-\sqrt{29}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{29}}{2*1}=\frac{3+\sqrt{29}}{2}
$