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X+X2=40
We move all terms to the left:
X+X2-(40)=0
We add all the numbers together, and all the variables
X^2+X-40=0
a = 1; b = 1; c = -40;
Δ = b2-4ac
Δ = 12-4·1·(-40)
Δ = 161
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-\sqrt{161}}{2*1}=\frac{-1-\sqrt{161}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{161}}{2*1}=\frac{-1+\sqrt{161}}{2} $
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