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3u^2-2u-21=0
a = 3; b = -2; c = -21;
Δ = b2-4ac
Δ = -22-4·3·(-21)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{256}=16$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-16}{2*3}=\frac{-14}{6} =-2+1/3 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+16}{2*3}=\frac{18}{6} =3 $
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