F(x)=3x^2+16x-12

Solution for F(x)=3x^2+16x-12 equation:

(F)=3F^2+16F-12

We move all terms to the left:

(F)-(3F^2+16F-12)=0

We get rid of parentheses

-3F^2+F-16F+12=0

We add all the numbers together, and all the variables

-3F^2-15F+12=0

a = -3; b = -15; c = +12;
Δ = b2-4ac
Δ = -152-4·(-3)·12
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{41}}{2*-3}=\frac{15-3\sqrt{41}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{41}}{2*-3}=\frac{15+3\sqrt{41}}{-6}
$