F(x)=3x^2-15x-42

Solution for F(x)=3x^2-15x-42 equation:

(F)=3F^2-15F-42

We move all terms to the left:

(F)-(3F^2-15F-42)=0

We get rid of parentheses

-3F^2+F+15F+42=0

We add all the numbers together, and all the variables

-3F^2+16F+42=0

a = -3; b = 16; c = +42;
Δ = b2-4ac
Δ = 162-4·(-3)·42
Δ = 760
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:

$\sqrt{\Delta}=\sqrt{760}=\sqrt{4*190}=\sqrt{4}*\sqrt{190}=2\sqrt{190}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{190}}{2*-3}=\frac{-16-2\sqrt{190}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{190}}{2*-3}=\frac{-16+2\sqrt{190}}{-6}
$