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X^2+40X+20=0
a = 1; b = 40; c = +20;
Δ = b2-4ac
Δ = 402-4·1·20
Δ = 1520
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1520}=\sqrt{16*95}=\sqrt{16}*\sqrt{95}=4\sqrt{95}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{95}}{2*1}=\frac{-40-4\sqrt{95}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{95}}{2*1}=\frac{-40+4\sqrt{95}}{2} $
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