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v^2+32v-185=0
a = 1; b = 32; c = -185;
Δ = b2-4ac
Δ = 322-4·1·(-185)
Δ = 1764
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{1764}=42$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-42}{2*1}=\frac{-74}{2} =-37 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+42}{2*1}=\frac{10}{2} =5 $
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