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X^2+40X+108=0
a = 1; b = 40; c = +108;
Δ = b2-4ac
Δ = 402-4·1·108
Δ = 1168
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1168}=\sqrt{16*73}=\sqrt{16}*\sqrt{73}=4\sqrt{73}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{73}}{2*1}=\frac{-40-4\sqrt{73}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{73}}{2*1}=\frac{-40+4\sqrt{73}}{2} $
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