Solve Quadratic Equations(Quadratic Formula)

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**Answer**

$x^{2}-10x+21=0$

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 $c=21$

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

$x=\frac{10\pm \sqrt{100-84}}{2}$

or, $x = \frac{10\pm \sqrt{16}}{2}$

or, $x = \frac{10\pm 4}{2}$

or, $x = \frac{14}{2}$ and $x = \frac{6}{2}$

or, $x=7$ and $x =3$

There are two distinct solutions given by $x = 7$ and $x = 3$