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

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

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

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-5F^2+4F+7=0

a = -5; b = 4; c = +7;
Δ = b2-4ac
Δ = 42-4·(-5)·7
Δ = 156
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:

$\sqrt{\Delta}=\sqrt{156}=\sqrt{4*39}=\sqrt{4}*\sqrt{39}=2\sqrt{39}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{39}}{2*-5}=\frac{-4-2\sqrt{39}}{-10}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{39}}{2*-5}=\frac{-4+2\sqrt{39}}{-10}
$