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.05v^2+2.2v-330=0
a = .05; b = 2.2; c = -330;
Δ = b2-4ac
Δ = 2.22-4·.05·(-330)
Δ = 70.84
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}$$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.2)-\sqrt{70.84}}{2*.05}=\frac{-2.2-\sqrt{70.84}}{0.1} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.2)+\sqrt{70.84}}{2*.05}=\frac{-2.2+\sqrt{70.84}}{0.1} $
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