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x^2+40x+184=0
a = 1; b = 40; c = +184;
Δ = b2-4ac
Δ = 402-4·1·184
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-12\sqrt{6}}{2*1}=\frac{-40-12\sqrt{6}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+12\sqrt{6}}{2*1}=\frac{-40+12\sqrt{6}}{2} $
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