# How many zeroes does the function have .25x^2-7x-12=y-4?

Aug 7, 2015

All real quadratic functions have at most two zeroes.

For this one, if you multiply by whole thing by $4$ and set $y = 0$:

${x}^{2} - 28 x - 48 = - 16$

$0 = {x}^{2} - 28 x - 32$

$x = \frac{- \left(- 28\right) \pm \sqrt{{\left(- 28\right)}^{2} - 4 \left(1\right) \left(- 32\right)}}{2 \left(1\right)}$

$= \frac{28 \pm \sqrt{784 + 128}}{2}$

$= \frac{28 \pm \sqrt{912}}{2} = \frac{28 \pm \sqrt{57 \cdot 4 \cdot 4}}{2}$

$= \frac{28 \pm 4 \sqrt{57}}{2}$

$= 14 \pm 2 \sqrt{57}$

Thus the two roots are:

$\textcolor{b l u e}{x = 14 + 2 \sqrt{57}}$
$\textcolor{b l u e}{x = 14 - 2 \sqrt{57}}$