# How do you find the roots, real and imaginary, of y=(x-5)(x-2) using the quadratic formula?

Mar 25, 2017

${x}_{1} = 5 + 0 i$

${x}_{2} = 2 + 0 i$

#### Explanation:

$y = \left(x - 5\right) \left(x - 2\right) = {x}^{2} - 7 x + 10$

Let $y = 0$

${x}^{2} - 7 x + 10 = 0$

Using Descartes' rule for $f \left(x\right)$, we know that there are two positive roots. Using it for $f \left(- x\right)$, we know that there can be zero or two negative roots.

$\Delta = {\left(- 7\right)}^{2} - 4 \left(10\right) \left(1\right) = 49 - 40 = 9$

$x = \frac{7 \pm \sqrt{9}}{2} = \frac{7 \pm 3}{2} = 5 , 2$

There are no negative roots and a quadratic can only have, at most, two real roots.

So the real values of $x$ are $5$ and $2$. These are also the complex roots are we can write them as $5 + 0 i$ and $2 + 0 i$.