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2v^2+11v-40=0
a = 2; b = 11; c = -40;
Δ = b2-4ac
Δ = 112-4·2·(-40)
Δ = 441
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{441}=21$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*2}=\frac{-32}{4} =-8 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*2}=\frac{10}{4} =2+1/2 $
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