# How do you use the quadratic formula to find both solutions to the quadratic equation 17x^2 = 12x ?

Jul 7, 2015

You don't have to use the quadratic formula.

#### Explanation:

One solution is in plain view: if $x = 0$ it holds.

If $x \ne 0$ we may divide by $x$:
$17 x = 12 \to x = \frac{12}{17} =$the other solution.

Jul 7, 2015

Using the quadratic formula is, perhaps, not the easiest way to solve this. (I say "perhaps", because easiness is not something someone else can decide for you.)

#### Explanation:

If you have been told to use the formula. Or if you are curious about how it would work. (After all, we tell students that every quadratic equation can be solved by using the formula.) Then here's how it goes:

Solve
$17 {x}^{2} = 12 x$

We need to put it in standard form for a quadratic equation:

$17 {x}^{2} - 12 x = 0$

Now we'll use $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this equation, we have:
$a = 17$
$b = - 12$
$c = 0$ (That's the one you may have had trouble with on your own.)

So we get:

$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(17\right) \left(0\right)}}{2 \left(17\right)}$

$x = \frac{12 \pm \sqrt{144 - 0}}{34}$

$= \frac{12 \pm 12}{34}$

So, we have:
$x = \frac{12 + 12}{34} = \frac{24}{34} = \frac{12}{17}$
Or
$x = \frac{12 - 12}{34} = \frac{0}{34} = 0$