F(1)=-5x^2+40x+73

Solution for F(1)=-5x^2+40x+73 equation:

(1)=-5F^2+40F+73

We move all terms to the left:

(1)-(-5F^2+40F+73)=0

We get rid of parentheses

5F^2-40F-73+1=0

We add all the numbers together, and all the variables

5F^2-40F-72=0

a = 5; b = -40; c = -72;
Δ = b2-4ac
Δ = -402-4·5·(-72)
Δ = 3040
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:

$\sqrt{\Delta}=\sqrt{3040}=\sqrt{16*190}=\sqrt{16}*\sqrt{190}=4\sqrt{190}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-4\sqrt{190}}{2*5}=\frac{40-4\sqrt{190}}{10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+4\sqrt{190}}{2*5}=\frac{40+4\sqrt{190}}{10}
$