F=5+-1x+-0.5x^2

Solution for F=5+-1x+-0.5x^2 equation:

=5+-1F+-0.5F^2

We move all terms to the left:

-(5+-1F+-0.5F^2)=0

We use the square of the difference formula

-(5-1F-0.5F^2)=0

We get rid of parentheses

0.5F^2+1F-5=0

We add all the numbers together, and all the variables

0.5F^2+F-5=0

a = 0.5; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·0.5·(-5)
Δ = 11
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{11}}{2*0.5}=\frac{-1-\sqrt{11}}{1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{11}}{2*0.5}=\frac{-1+\sqrt{11}}{1}
$