F(x)=x2+3x+-1

Solution for F(x)=x2+3x+-1 equation:

(F)=F2+3F+-1

We move all terms to the left:

(F)-(F2+3F+-1)=0

We add all the numbers together, and all the variables

-(+F^2+3F+-1)+F=0

We use the square of the difference formula

-(+F^2+3F-1)+F=0

We get rid of parentheses

-F^2-3F+F+1=0

We add all the numbers together, and all the variables

-1F^2-2F+1=0

a = -1; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-1)·1
Δ = 8
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:

$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{2}}{2*-1}=\frac{2-2\sqrt{2}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{2}}{2*-1}=\frac{2+2\sqrt{2}}{-2}
$