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u^2+3u=2
We move all terms to the left:
u^2+3u-(2)=0
a = 1; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·1·(-2)
Δ = 17
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}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{17}}{2*1}=\frac{-3-\sqrt{17}}{2} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{17}}{2*1}=\frac{-3+\sqrt{17}}{2} $
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