We have learned a couple of methods for solving quadratic equations algebraically:

• By rewriting the equation as so that one side is 0 and the other side is in factored form, then using the zero product property
• By completing the square

Some equations can be solved quickly with one of these methods, but many cannot. Here is an example: $5x^2-3x-1=0$. The expression on the left cannot be rewritten in factored form with rational coefficients. Because the coefficient of the squared term is not a perfect square, and the coefficient of the linear term is an odd number, completing the square would be inconvenient and would result in a perfect square with fractions.

The quadratic formula can be used to find the solutions to any quadratic equation, including those that are tricky to solve with other methods.

For the equation $5x^2-3x-1=0$, we see that $a=5$, $b=\text-3$, and $c=\text-1$. Let’s solve it!

$\displaystyle \begin{aligned} x &=\dfrac{\text- b \pm \sqrt{b^2-4ac}}{2a} &\qquad &\text{the quadratic formula}\\ x &=\dfrac{\text-(\text-3) \pm \sqrt{(\text-3)^2-4(5)(\text-1)}}{2(5)} &\qquad &\text{Substitute the values of }a, b, \text{and }c. &\\ x &=\dfrac{3 \pm \sqrt{9+20}}{10} &\qquad &\text {Evaluate each part of the expression.} \\ x &=\dfrac{3 \pm \sqrt{29}}{10}\end{aligned}$

A calculator gives approximate solutions of 0.84 and -0.24 for $\frac{3 + \sqrt{29}}{10}$ and $\frac{3 - \sqrt{29}}{10}$.

We can also use the formula for simpler equations like $x^2-9x+8=0$, but it may not be the most efficient way. If the quadratic expression can be easily rewritten in factored form or made into a perfect square, those methods may be preferable. For example, rewriting $x^2-9x+8=0$ as $(x-1)(x-8)=0$ immediately tells us that the solutions are 1 and 8.