F(x)=3x^2+x-16

Solution for F(x)=3x^2+x-16 equation:

(F)=3F^2+F-16

We move all terms to the left:

(F)-(3F^2+F-16)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-3F^2+16=0

a = -3; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-3)·16
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{3}}{2*-3}=\frac{0-8\sqrt{3}}{-6}
=-\frac{8\sqrt{3}}{-6}
=-\frac{4\sqrt{3}}{-3}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{3}}{2*-3}=\frac{0+8\sqrt{3}}{-6}
=\frac{8\sqrt{3}}{-6}
=\frac{4\sqrt{3}}{-3}
$