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0.5v^2+v-80=0
a = 0.5; b = 1; c = -80;
Δ = b2-4ac
Δ = 12-4·0.5·(-80)
Δ = 161
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{-(1)-\sqrt{161}}{2*0.5}=\frac{-1-\sqrt{161}}{1} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*0.5}=\frac{-1+\sqrt{161}}{1} $
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