F(x)=5-(4x^2)

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

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

We move all terms to the left:

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

determiningTheFunctionDomain
-(5-4F^2)+F=0

We get rid of parentheses

4F^2+F-5=0

a = 4; b = 1; c = -5;
Δ = b2-4ac
Δ = 12-4·4·(-5)
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$F_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-9}{2*4}=\frac{-10}{8}
=-1+1/4
$
$F_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+9}{2*4}=\frac{8}{8}
=1
$