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v^2-17v+42=0
a = 1; b = -17; c = +42;
Δ = b2-4ac
Δ = -172-4·1·42
Δ = 121
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{121}=11$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-11}{2*1}=\frac{6}{2} =3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+11}{2*1}=\frac{28}{2} =14 $
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