Solve Quadratic Equations(Quadratic Formula)

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**Answer**

$x^{2}-5x+4=0$

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*, $a=1$, $b=-5$, and $c=4$

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

$x=\frac{5\pm \sqrt{25-16}}{2}$

or, $x = \frac{5\pm \sqrt{9}}{2}$

or, $x = \frac{5\pm 3}{2}$

or, $x = \frac{8}{2}$ and $x = \frac{2}{2}$

or, $x=4$ and $x =1$

There are two distinct solutions given by $x = 4$ and $x = 1$