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3u^2-3u-90=0
a = 3; b = -3; c = -90;
Δ = b2-4ac
Δ = -32-4·3·(-90)
Δ = 1089
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{1089}=33$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-33}{2*3}=\frac{-30}{6} =-5 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+33}{2*3}=\frac{36}{6} =6 $
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