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15v^2+50v-40=0
a = 15; b = 50; c = -40;
Δ = b2-4ac
Δ = 502-4·15·(-40)
Δ = 4900
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{4900}=70$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-70}{2*15}=\frac{-120}{30} =-4 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+70}{2*15}=\frac{20}{30} =2/3 $
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