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u^2+4u-33=0
a = 1; b = 4; c = -33;
Δ = b2-4ac
Δ = 42-4·1·(-33)
Δ = 148
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}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{37}}{2*1}=\frac{-4-2\sqrt{37}}{2} $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{37}}{2*1}=\frac{-4+2\sqrt{37}}{2} $
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