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3v^2+20v+17=0
a = 3; b = 20; c = +17;
Δ = b2-4ac
Δ = 202-4·3·17
Δ = 196
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{196}=14$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-14}{2*3}=\frac{-34}{6} =-5+2/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+14}{2*3}=\frac{-6}{6} =-1 $
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