# How many solutions does -12x^2-4x+5=0 have?

Mar 5, 2018

Two

#### Explanation:

It can only have 2 or less solutions because the highst power of x is 2 ($- 12 {x}^{\textcolor{b l u e}{2}}$). Lets check if it has 2, 1 or no solutions:

-12x^2-4x+5=0|:(-12)
${x}^{2} + \frac{1}{3} x - \frac{5}{12} = 0$
$\textcolor{b l u e}{{x}^{2} + \frac{1}{3} x + \frac{1}{36}} \textcolor{red}{- \frac{1}{36} - \frac{5}{12}} = 0$
$\textcolor{b l u e}{{\left(x + \frac{1}{6}\right)}^{2}} \textcolor{red}{- \frac{16}{36}} = 0 | + \frac{16}{36}$
${\left(x + \frac{1}{6}\right)}^{2} = \frac{16}{36} | \sqrt{}$
$x + \frac{1}{6} = \pm \frac{2}{3} | - \frac{1}{6}$
$x = \pm \frac{2}{3} - \frac{1}{6}$
${x}_{1} = \frac{1}{2} \mathmr{and} {x}_{2} = - \frac{5}{6}$

Mar 5, 2018

Method shown below, you do the math.

#### Explanation:

Rewrite the equation, change signs on both side:

$12 {x}^{2} + 4 x - 5 = 0$
This can be seen to be the familar quadratic equation

$a {x}^{2} + b x + c$ with a solution:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute values to a, b, c to get the answer