# How do I use the quadratic formula to find the zeros of f?

*Given a function of the form $f \left(x\right) = a {x}^{2} + b x + c$, one can find the zeroes of the function (that is, where $f \left(x\right) = 0$) by using the quadratic formula: $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.
If the discriminant ${b}^{2} - 4 a c$ is less than zero, these roots will be complex or imaginary, and if the discriminant is greater than or equal to 0. these roots will be real.
The use of $\pm$ informs us that there are two solutions here; one where we are subtracting $\sqrt{{b}^{2} - 4 a c}$, and one where we are adding it.*
For a step by step example, assume the function $f \left(x\right) = {x}^{2} - 7 x + 10$. Here $a = 1 , b = - 7 , c = 10.$ If we set $f \left(x\right) = 0$, then the values of x for which $f \left(x\right) = 0$, that is to say the roots of $f \left(x\right)$, would be determined by the quadratic formula: $x = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \left(1\right) \left(10\right)}}{2 \left(1\right)} = \frac{7 \pm \sqrt{49 - 40}}{2} = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2}$. Thus, our roots will be at $x = \frac{7 + 3}{2}$ and $x = \frac{7 - 3}{2}$ or $x = 5$ and $x = 2$.
On the graph of the function, we would see that the parabola crosses the x-axis at $x = 2$ and $x = 5$.