7r2+42r=0

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Solution for 7r2+42r=0 equation: 



7r^2+42r=0

a = 7; b = 42; c = 0;
Δ = b2-4ac
Δ = 422-4·7·0
Δ = 1764
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1764}=42$


$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-42}{2*7}=\frac{-84}{14}
=-6
$


$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+42}{2*7}=\frac{0}{14}
=0
$

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