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2q^2-12q=0
a = 2; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·2·0
Δ = 144
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{144}=12$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*2}=\frac{0}{4} =0 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*2}=\frac{24}{4} =6 $
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