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 add all the numbers together, and all the variables

-(3F^2-4F-5)-4=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 = 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}
$