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

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

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

We move all terms to the left:

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

We get rid of parentheses

-3F^2-4F-5+4=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 = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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