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3u^2+27u-45=0
a = 3; b = 27; c = -45;
Δ = b2-4ac
Δ = 272-4·3·(-45)
Δ = 1269
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}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1269}=\sqrt{9*141}=\sqrt{9}*\sqrt{141}=3\sqrt{141}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-3\sqrt{141}}{2*3}=\frac{-27-3\sqrt{141}}{6} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+3\sqrt{141}}{2*3}=\frac{-27+3\sqrt{141}}{6} $
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