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v(2v+5)=0
We multiply parentheses
2v^2+5v=0
a = 2; b = 5; c = 0;
Δ = b2-4ac
Δ = 52-4·2·0
Δ = 25
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{25}=5$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-5}{2*2}=\frac{-10}{4} =-2+1/2 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+5}{2*2}=\frac{0}{4} =0 $
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