f(2)=6,f(-1)=-3

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Solution for f(2)=6,f(-1)=-3 equation:



f(2)=6.f(-1)=-3
We move all terms to the left:
f(2)-(6.f(-1))=0
We add all the numbers together, and all the variables
f^2-(6.f(-1))=0
We calculate terms in parentheses: -(6.f(-1)), so:
6.f(-1)
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 $

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