f(-21)=2/3f+9

Solution for f(-21)=2/3f+9 equation:

f(-21)=2/3f+9

We move all terms to the left:

f(-21)-(2/3f+9)=0


Domain of the equation: 3f+9)!=0

f∈R


We multiply parentheses

-21f-(2/3f+9)=0

We get rid of parentheses

-21f-2/3f-9=0

We multiply all the terms by the denominator


-21f*3f-9*3f-2=0

Wy multiply elements

-63f^2-27f-2=0

a = -63; b = -27; c = -2;
Δ = b2-4ac
Δ = -272-4·(-63)·(-2)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-27)-15}{2*-63}=\frac{12}{-126}
=-2/21
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-27)+15}{2*-63}=\frac{42}{-126}
=-1/3
$