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2x^2-9x-65=x^-x
We move all terms to the left:
2x^2-9x-65-(x^-x)=0
We get rid of parentheses
2x^2-9x-x^+x-65=0
We add all the numbers together, and all the variables
2x^2-9x-65=0
a = 2; b = -9; c = -65;
Δ = b2-4ac
Δ = -92-4·2·(-65)
Δ = 601
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{-(-9)-\sqrt{601}}{2*2}=\frac{9-\sqrt{601}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{601}}{2*2}=\frac{9+\sqrt{601}}{4} $
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