-3b+b^2=0

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Solution for -3b+b^2=0 equation: 



-3b+b^2=0

a = 1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·1·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*1}=\frac{0}{2}
=0
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*1}=\frac{6}{2}
=3
$

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