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-3v^2-5v-2=0
a = -3; b = -5; c = -2;
Δ = b2-4ac
Δ = -52-4·(-3)·(-2)
Δ = 1
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{1}=1$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-1}{2*-3}=\frac{4}{-6} =-2/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*-3}=\frac{6}{-6} =-1 $
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