b^2+81b=0

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Solution for b^2+81b=0 equation: 



b^2+81b=0

a = 1; b = 81; c = 0;
Δ = b2-4ac
Δ = 812-4·1·0
Δ = 6561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6561}=81$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(81)-81}{2*1}=\frac{-162}{2}
=-81
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(81)+81}{2*1}=\frac{0}{2}
=0
$

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