F(1)=-3x^2+2x+5

Solution for F(1)=-3x^2+2x+5 equation:

(1)=-3F^2+2F+5

We move all terms to the left:

(1)-(-3F^2+2F+5)=0

We get rid of parentheses

3F^2-2F-5+1=0

We add all the numbers together, and all the variables

3F^2-2F-4=0

a = 3; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·3·(-4)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{13}}{2*3}=\frac{2-2\sqrt{13}}{6}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{13}}{2*3}=\frac{2+2\sqrt{13}}{6}
$