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2v^2+19v+42=0
a = 2; b = 19; c = +42;
Δ = b2-4ac
Δ = 192-4·2·42
Δ = 25
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{25}=5$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-5}{2*2}=\frac{-24}{4} =-6 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+5}{2*2}=\frac{-14}{4} =-3+1/2 $
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