F(6)=3(x)^2+4x+5

Solution for F(6)=3(x)^2+4x+5 equation:

(6)=3(F)^2+4F+5

We move all terms to the left:

(6)-(3(F)^2+4F+5)=0

We get rid of parentheses

-3F^2-4F-5+6=0

We add all the numbers together, and all the variables

-3F^2-4F+1=0

a = -3; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·(-3)·1
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{7}}{2*-3}=\frac{4-2\sqrt{7}}{-6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{7}}{2*-3}=\frac{4+2\sqrt{7}}{-6}
$