F(15)=2x^2+3

Solution for F(15)=2x^2+3 equation:

(15)=2F^2+3

We move all terms to the left:

(15)-(2F^2+3)=0

We get rid of parentheses

-2F^2-3+15=0

We add all the numbers together, and all the variables

-2F^2+12=0

a = -2; b = 0; c = +12;
Δ = b2-4ac
Δ = 02-4·(-2)·12
Δ = 96
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:

$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{6}}{2*-2}=\frac{0-4\sqrt{6}}{-4}
=-\frac{4\sqrt{6}}{-4}
=-\frac{\sqrt{6}}{-1}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{6}}{2*-2}=\frac{0+4\sqrt{6}}{-4}
=\frac{4\sqrt{6}}{-4}
=\frac{\sqrt{6}}{-1}
$