Solve Quadratic Equations(Completing the Square)

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**Answer**

$x^{2}-13x+36=0$

or, $x^{2}-13x+36-36=-36$

or, $x^{2}-13x=-36$

or, $x^{2}-13x+\frac{169}{4}=-36+\frac{169}{4}$

or, $x^{2}-13x+\frac{169}{4}=\frac{25}{4}$

or, $\left(x-\frac{13}{2}\right)^2=\left(\frac{5}{2}\right)^2$

or, $x-\frac{13}{2}=\pm\frac{5}{2}$

or, $x =\pm\frac{5}{2}-\frac{13}{2}$

or, $x=-4$ and $x=-9$

There are two distinct solutions given by $x = -4$ and $x = -9$