f(-2)=(1/3)^-2

Solution for f(-2)=(1/3)^-2 equation:

f(-2)=(1/3)^-2

We move all terms to the left:

f(-2)-((1/3)^-2)=0

We add all the numbers together, and all the variables

f(-2)-((+1/3)^-2)=0

We multiply parentheses

-2f-((+1/3)^-2)=0

We multiply all the terms by the denominator


-2f*3)^-2)-((+1=0

Wy multiply elements

-6f^2+1=0

a = -6; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-6)·1
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*-6}=\frac{0-2\sqrt{6}}{-12}
=-\frac{2\sqrt{6}}{-12}
=-\frac{\sqrt{6}}{-6}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*-6}=\frac{0+2\sqrt{6}}{-12}
=\frac{2\sqrt{6}}{-12}
=\frac{\sqrt{6}}{-6}
$