Solve Quadratic Equations(Quadratic Formula)

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**Answer**

$x^{2}-10x+25=0$

The solutions of a *quadratic equation* $ax^2+bx + c = 0$ are given by$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

In this *quadratic equation*, $a=1$, $b=-10$, and $c=25$

Substituting these values of $a$, $b$ and $c$ in the above formula, we get

$x=\frac{10\pm \sqrt{100-100}}{2}$

or, $x = \frac{10\pm \sqrt{0}}{2}$

or, $x = \frac{10\pm 0}{2}$

or, $x = \frac{10}{2}$ and $x = \frac{10}{2}$

or, $x=5$ and $x =5$

There is only one unique solution given by $x=5$ of multiplicity $2$.