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X^2+20X-360=0
a = 1; b = 20; c = -360;
Δ = b2-4ac
Δ = 202-4·1·(-360)
Δ = 1840
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1840}=\sqrt{16*115}=\sqrt{16}*\sqrt{115}=4\sqrt{115}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-4\sqrt{115}}{2*1}=\frac{-20-4\sqrt{115}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+4\sqrt{115}}{2*1}=\frac{-20+4\sqrt{115}}{2} $
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