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5v^2+3v-110=0
a = 5; b = 3; c = -110;
Δ = b2-4ac
Δ = 32-4·5·(-110)
Δ = 2209
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{2209}=47$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-47}{2*5}=\frac{-50}{10} =-5 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+47}{2*5}=\frac{44}{10} =4+2/5 $
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