f^2+13f=0

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Solution for f^2+13f=0 equation: 



f^2+13f=0

a = 1; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·1·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$


$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*1}=\frac{-26}{2}
=-13
$


$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*1}=\frac{0}{2}
=0
$

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