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X^2+24X+112=0
a = 1; b = 24; c = +112;
Δ = b2-4ac
Δ = 242-4·1·112
Δ = 128
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-8\sqrt{2}}{2*1}=\frac{-24-8\sqrt{2}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+8\sqrt{2}}{2*1}=\frac{-24+8\sqrt{2}}{2} $
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