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x+x2-5=360
We move all terms to the left:
x+x2-5-(360)=0
We add all the numbers together, and all the variables
x^2+x-365=0
a = 1; b = 1; c = -365;
Δ = b2-4ac
Δ = 12-4·1·(-365)
Δ = 1461
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{1461}}{2*1}=\frac{-1-\sqrt{1461}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{1461}}{2*1}=\frac{-1+\sqrt{1461}}{2} $
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