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v^2-3v-20=0
a = 1; b = -3; c = -20;
Δ = b2-4ac
Δ = -32-4·1·(-20)
Δ = 89
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{-(-3)-\sqrt{89}}{2*1}=\frac{3-\sqrt{89}}{2} $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{89}}{2*1}=\frac{3+\sqrt{89}}{2} $
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