F(x)=10x^2+23x-42

Solution for F(x)=10x^2+23x-42 equation:

(F)=10F^2+23F-42

We move all terms to the left:

(F)-(10F^2+23F-42)=0

We get rid of parentheses

-10F^2+F-23F+42=0

We add all the numbers together, and all the variables

-10F^2-22F+42=0

a = -10; b = -22; c = +42;
Δ = b2-4ac
Δ = -222-4·(-10)·42
Δ = 2164
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:

$\sqrt{\Delta}=\sqrt{2164}=\sqrt{4*541}=\sqrt{4}*\sqrt{541}=2\sqrt{541}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{541}}{2*-10}=\frac{22-2\sqrt{541}}{-20}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{541}}{2*-10}=\frac{22+2\sqrt{541}}{-20}
$