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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3F^2+4F-5=0

a = 3; b = 4; c = -5;
Δ = b2-4ac
Δ = 42-4·3·(-5)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{19}}{2*3}=\frac{-4-2\sqrt{19}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{19}}{2*3}=\frac{-4+2\sqrt{19}}{6}
$