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5v^2+34v-48=0
a = 5; b = 34; c = -48;
Δ = b2-4ac
Δ = 342-4·5·(-48)
Δ = 2116
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{2116}=46$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-46}{2*5}=\frac{-80}{10} =-8 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+46}{2*5}=\frac{12}{10} =1+1/5 $
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