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X^2+40X+345=0
a = 1; b = 40; c = +345;
Δ = b2-4ac
Δ = 402-4·1·345
Δ = 220
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-2\sqrt{55}}{2*1}=\frac{-40-2\sqrt{55}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+2\sqrt{55}}{2*1}=\frac{-40+2\sqrt{55}}{2} $
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