14r2+7r=0

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Solution for 14r2+7r=0 equation: 



14r^2+7r=0

a = 14; b = 7; c = 0;
Δ = b2-4ac
Δ = 72-4·14·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-7}{2*14}=\frac{-14}{28}
=-1/2
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+7}{2*14}=\frac{0}{28}
=0
$

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