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3v^2=-3v
We move all terms to the left:
3v^2-(-3v)=0
We get rid of parentheses
3v^2+3v=0
a = 3; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·3·0
Δ = 9
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{9}=3$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*3}=\frac{-6}{6} =-1 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*3}=\frac{0}{6} =0 $
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