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18q-2q^2=0
a = -2; b = 18; c = 0;
Δ = b2-4ac
Δ = 182-4·(-2)·0
Δ = 324
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{324}=18$$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-18}{2*-2}=\frac{-36}{-4} =+9 $$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+18}{2*-2}=\frac{0}{-4} =0 $
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