F(x)=3x2+12x+5

Solution for F(x)=3x2+12x+5 equation:

(F)=3F^2+12F+5

We move all terms to the left:

(F)-(3F^2+12F+5)=0

We get rid of parentheses

-3F^2+F-12F-5=0

We add all the numbers together, and all the variables

-3F^2-11F-5=0

a = -3; b = -11; c = -5;
Δ = b2-4ac
Δ = -112-4·(-3)·(-5)
Δ = 61
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-\sqrt{61}}{2*-3}=\frac{11-\sqrt{61}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+\sqrt{61}}{2*-3}=\frac{11+\sqrt{61}}{-6}
$