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-3v^2+63v=0
a = -3; b = 63; c = 0;
Δ = b2-4ac
Δ = 632-4·(-3)·0
Δ = 3969
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{3969}=63$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(63)-63}{2*-3}=\frac{-126}{-6} =+21 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(63)+63}{2*-3}=\frac{0}{-6} =0 $
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