-7b^2+13b=0

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Solution for -7b^2+13b=0 equation: 



-7b^2+13b=0

a = -7; b = 13; c = 0;
Δ = b2-4ac
Δ = 132-4·(-7)·0
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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


$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-13}{2*-7}=\frac{-26}{-14}
=1+6/7
$


$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+13}{2*-7}=\frac{0}{-14}
=0
$

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