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2x+x^2=40
We move all terms to the left:
2x+x^2-(40)=0
a = 1; b = 2; c = -40;
Δ = b2-4ac
Δ = 22-4·1·(-40)
Δ = 164
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{41}}{2*1}=\frac{-2-2\sqrt{41}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{41}}{2*1}=\frac{-2+2\sqrt{41}}{2} $
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