r^2-15r=0

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Solution for r^2-15r=0 equation:



r^2-15r=0
a = 1; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·1·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*1}=\frac{0}{2} =0 $
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*1}=\frac{30}{2} =15 $

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