# How do you find the zeros, real and imaginary, of y=x^2 -34x+1 using the quadratic formula?

Nov 16, 2017

See a solution process below:

#### Explanation:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 34}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{1}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{- 34} \pm \sqrt{{\textcolor{b l u e}{- 34}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{1}}$

$x = \frac{34 \pm \sqrt{1156 - 4}}{2}$

$x = \frac{34 \pm \sqrt{1152}}{2}$

$x = \frac{34 - \sqrt{576 \cdot 2}}{2}$ and $x = \frac{34 + \sqrt{576 \cdot 2}}{2}$

$x = \frac{34 - \sqrt{576} \sqrt{2}}{2}$ and $x = \frac{34 + \sqrt{576} \sqrt{2}}{2}$

$x = \frac{34 - 24 \sqrt{2}}{2}$ and $x = \frac{34 + 24 \sqrt{2}}{2}$

$x = 17 - 12 \sqrt{2}$ and $x = 17 + 12 \sqrt{2}$