# How do you solve x^2 - 12x + 52 = 0?

Jun 8, 2015

${x}^{2} - 12 x + 52 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:
$a = 1 , b = - 12 , c = 52$

The Discriminant is given by:
$\Delta = {b}^{2} - 4 \cdot a \cdot c$
$= {\left(- 12\right)}^{2} - \left(4 \cdot 1 \cdot 52\right)$
$= 144 - 208$
$= - 64$

If $\Delta = 0$ then there is only one solution.
(for $\Delta > 0$ there are two solutions,
for $\Delta < 0$ there are no real solutions)

As $\Delta = - 64$, this equation has NO REAL SOLUTIONS

Note :
The solutions are normally found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

As $\Delta = - 64$, $x = \frac{- \left(- 12\right) \pm \sqrt{- 64}}{2 \cdot 1} = \frac{12 \pm \sqrt{- 64}}{2}$

Jun 8, 2015

${x}^{2} - 12 x + 52$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = - 12$ and $c = 52$. This has discriminant given by the formula:

$\Delta = {b}^{2} - 4 a c = {\left(- 12\right)}^{2} - \left(4 \times 1 \times 52\right)$

$= 144 - 208 = - 64 < 0$

So ${x}^{2} - 12 x + 52 = 0$ has no real roots. It has two distinct complex roots.