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3v^2-25v+28=0
a = 3; b = -25; c = +28;
Δ = b2-4ac
Δ = -252-4·3·28
Δ = 289
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{289}=17$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-17}{2*3}=\frac{8}{6} =1+1/3 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+17}{2*3}=\frac{42}{6} =7 $
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