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(5v+2)(8v)=0
We multiply parentheses
40v^2+16v=0
a = 40; b = 16; c = 0;
Δ = b2-4ac
Δ = 162-4·40·0
Δ = 256
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{256}=16$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-16}{2*40}=\frac{-32}{80} =-2/5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+16}{2*40}=\frac{0}{80} =0 $
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