# How do you solve 6b ^ { 2} - b + 1= 0?

Mar 20, 2018

See a solution process below:

#### Explanation:

We can use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{6}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{1}$ for $\textcolor{g r e e n}{c}$ gives:

$b = \frac{- \textcolor{b l u e}{- 1} \pm \sqrt{{\textcolor{b l u e}{- 1}}^{2} - \left(4 \cdot \textcolor{red}{6} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{6}}$

$b = \frac{\textcolor{b l u e}{1} \pm \sqrt{1 - 24}}{12}$

$b = \frac{\textcolor{b l u e}{1} \pm \sqrt{- 23}}{12}$