7q^2+28q=0

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Solution for 7q^2+28q=0 equation: 



7q^2+28q=0

a = 7; b = 28; c = 0;
Δ = b2-4ac
Δ = 282-4·7·0
Δ = 784
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{784}=28$


$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-28}{2*7}=\frac{-56}{14}
=-4
$


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

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