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3u^2+25u-18=0
a = 3; b = 25; c = -18;
Δ = b2-4ac
Δ = 252-4·3·(-18)
Δ = 841
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{841}=29$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-29}{2*3}=\frac{-54}{6} =-9 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+29}{2*3}=\frac{4}{6} =2/3 $
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