q(q-7)=0

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Solution for q(q-7)=0 equation:



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

$\sqrt{\Delta}=\sqrt{49}=7$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*1}=\frac{0}{2} =0 $
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*1}=\frac{14}{2} =7 $

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