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x(squared)-x-5= 0 x=?
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use the quadratic formula
Your equation is in form \[ax^2+bx+c=0\] your equation is\[x^2-x-5=0\] or\[1\times x^2-1 \times x -5=0\] So you can see that coefficients for your equation is a=1, b=-1, c=-5 then you can find the roots using this formula: \[x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1)\pm \sqrt{{(-1)^2}-4\times 1 \times (-5)}}{2\times 1}=\frac{1\pm \sqrt{21}}{2}\] \[x_{1}=\frac{1+\sqrt{21}}{2}\] \[x_2=\frac{1-\sqrt{21}}{2}\]
thank you, I think I forgot to set these over 2!
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