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X^2+20X+20X+100=0
We add all the numbers together, and all the variables
X^2+40X+100=0
a = 1; b = 40; c = +100;
Δ = b2-4ac
Δ = 402-4·1·100
Δ = 1200
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1200}=\sqrt{400*3}=\sqrt{400}*\sqrt{3}=20\sqrt{3}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-20\sqrt{3}}{2*1}=\frac{-40-20\sqrt{3}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+20\sqrt{3}}{2*1}=\frac{-40+20\sqrt{3}}{2} $
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