F(x)=-x^2+.03+5

Solution for F(x)=-x^2+.03+5 equation:

(F)=-F^2+.03+5

We move all terms to the left:

(F)-(-F^2+.03+5)=0

We get rid of parentheses

F^2+F-.03-5=0

We add all the numbers together, and all the variables

F^2+F-5.03=0

a = 1; b = 1; c = -5.03;
Δ = b2-4ac
Δ = 12-4·1·(-5.03)
Δ = 21.12
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{-(1)-\sqrt{21.12}}{2*1}=\frac{-1-\sqrt{21.12}}{2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{21.12}}{2*1}=\frac{-1+\sqrt{21.12}}{2}
$