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v^2-9v-20=0
a = 1; b = -9; c = -20;
Δ = b2-4ac
Δ = -92-4·1·(-20)
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{-(-9)-\sqrt{161}}{2*1}=\frac{9-\sqrt{161}}{2} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{161}}{2*1}=\frac{9+\sqrt{161}}{2} $
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