F(x)=x^2+3x-19

Solution for F(x)=x^2+3x-19 equation:

(F)=F^2+3F-19

We move all terms to the left:

(F)-(F^2+3F-19)=0

We get rid of parentheses

-F^2+F-3F+19=0

We add all the numbers together, and all the variables

-1F^2-2F+19=0

a = -1; b = -2; c = +19;
Δ = b2-4ac
Δ = -22-4·(-1)·19
Δ = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-4\sqrt{5}}{2*-1}=\frac{2-4\sqrt{5}}{-2}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+4\sqrt{5}}{2*-1}=\frac{2+4\sqrt{5}}{-2}
$