# How do you solve 3b^2-6b+51=0?

Dec 29, 2016

There are no Real solutions;
as Complex solutions: $b = 1 \pm 4 i$

#### Explanation:

Given $3 {b}^{2} - 6 b + 51 = 0$

We could apply the quadratic formula directly to obtain solutions,
but, noticing that all terms are divisible by $3$, we can reduce the complexity of calculations by replacing the given equation with
$\textcolor{w h i t e}{\text{XXX}} {b}^{2} - 2 b + 17 = 0$

The quadratic formula tells us that for an equation in the form:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$
the solution(s) are given by:
$\textcolor{w h i t e}{\text{XXX}} x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{\textcolor{b l u e}{b} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}}}{2 \textcolor{red}{a}}$

The equation we have been given has complicated this situation (probably intentionally) by using $b$ instead of $x$ as it's variable. This can cause some confusion with the constant $\textcolor{b l u e}{b}$ in the standard form of the quadratic formula.

Let's re-write the quadratic formula with $b$ as the variable and different letters as the place-holder variables:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{p} {b}^{2} + \textcolor{b l u e}{q} b + \textcolor{g r e e n}{r} = 0$
has solutions:
$\textcolor{w h i t e}{\text{XXX}} b = \frac{- \textcolor{b l u e}{q} \pm \sqrt{{\textcolor{b l u e}{q}}^{2} - 4 \textcolor{red}{p} \textcolor{g r e e n}{r}}}{2 \textcolor{red}{p}}$

Using ${b}^{2} \textcolor{b l u e}{- 2} b \textcolor{g r e e n}{+ 17} = 0$ as our equation
we have
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{p} = \textcolor{red}{1}$ (by default)
$\textcolor{w h i t e}{\text{XXX")color(blue)q=color(blue)(} \left(- 2\right)}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{r} = \textcolor{g r e e n}{17}$

So the solutions are
color(white)("XXX")b=(color(blue)2+-sqrt((color(blue)(""-2))^2-4 * color(red) 1 * color(green)(17)))/(2 * color(red)1)

$\textcolor{w h i t e}{\text{XXX}} = \frac{2 \pm \sqrt{- 64}}{2} = \frac{2 \pm 8 \sqrt{- 1}}{2} = 1 \pm 4 i$