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X^2+42X+184=0
a = 1; b = 42; c = +184;
Δ = b2-4ac
Δ = 422-4·1·184
Δ = 1028
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1028}=\sqrt{4*257}=\sqrt{4}*\sqrt{257}=2\sqrt{257}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{257}}{2*1}=\frac{-42-2\sqrt{257}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{257}}{2*1}=\frac{-42+2\sqrt{257}}{2} $
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