F(3)=5x^2-4

Solution for F(3)=5x^2-4 equation:

(3)=5F^2-4

We move all terms to the left:

(3)-(5F^2-4)=0

We get rid of parentheses

-5F^2+4+3=0

We add all the numbers together, and all the variables

-5F^2+7=0

a = -5; b = 0; c = +7;
Δ = b2-4ac
Δ = 02-4·(-5)·7
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{35}}{2*-5}=\frac{0-2\sqrt{35}}{-10}
=-\frac{2\sqrt{35}}{-10}
=-\frac{\sqrt{35}}{-5}
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{35}}{2*-5}=\frac{0+2\sqrt{35}}{-10}
=\frac{2\sqrt{35}}{-10}
=\frac{\sqrt{35}}{-5}
$