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3v^2+3v-168=0
a = 3; b = 3; c = -168;
Δ = b2-4ac
Δ = 32-4·3·(-168)
Δ = 2025
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{2025}=45$$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-45}{2*3}=\frac{-48}{6} =-8 $$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+45}{2*3}=\frac{42}{6} =7 $
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