F(x)=2x^2-16-35

Solution for F(x)=2x^2-16-35 equation:

(F)=2F^2-16-35

We move all terms to the left:

(F)-(2F^2-16-35)=0

We get rid of parentheses

-2F^2+F+16+35=0

We add all the numbers together, and all the variables

-2F^2+F+51=0

a = -2; b = 1; c = +51;
Δ = b2-4ac
Δ = 12-4·(-2)·51
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{409}}{2*-2}=\frac{-1-\sqrt{409}}{-4}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{409}}{2*-2}=\frac{-1+\sqrt{409}}{-4}
$