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18v^2-4v=0
a = 18; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·18·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*18}=\frac{0}{36} =0 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*18}=\frac{8}{36} =2/9 $
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