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3v^2=19v+14
We move all terms to the left:
3v^2-(19v+14)=0
We get rid of parentheses
3v^2-19v-14=0
a = 3; b = -19; c = -14;
Δ = b2-4ac
Δ = -192-4·3·(-14)
Δ = 529
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{529}=23$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-23}{2*3}=\frac{-4}{6} =-2/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+23}{2*3}=\frac{42}{6} =7 $
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