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3u=-2u^2+2
We move all terms to the left:
3u-(-2u^2+2)=0
We get rid of parentheses
2u^2+3u-2=0
a = 2; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·2·(-2)
Δ = 25
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{25}=5$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-5}{2*2}=\frac{-8}{4} =-2 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+5}{2*2}=\frac{2}{4} =1/2 $
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