f(2)=1f(-6)=65

Solution for f(2)=1f(-6)=65 equation:

f(2)=1f(-6)=65

We move all terms to the left:

f(2)-(1f(-6))=0

We add all the numbers together, and all the variables

f^2-(1f(-6))=0


We calculate terms in parentheses: -(1f(-6)), so:

1f(-6)

We multiply parentheses

-6f

Back to the equation:

-(-6f)


We get rid of parentheses

f^2+6f=0

a = 1; b = 6; c = 0;
Δ = b2-4ac
Δ = 62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6}{2*1}=\frac{-12}{2}
=-6
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6}{2*1}=\frac{0}{2}
=0
$