# How do you solve the quadratic using the quadratic formula given 14m^2+1=6m^2+7m?

Aug 3, 2017

See a solution process below:

#### Explanation:

First we need to convert the equation to standard form:

$- \textcolor{red}{6 {m}^{2}} + 14 {m}^{2} - \textcolor{b l u e}{7 m} + 1 = - \textcolor{red}{6 {m}^{2}} + 6 {m}^{2} + 7 m - \textcolor{b l u e}{7 m}$

$\left(- \textcolor{red}{6} + 14\right) {m}^{2} - 7 m + 1 = 0 + 0$

$8 {m}^{2} - 7 m + 1 = 0$

Now, 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:

$\textcolor{red}{8}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 7}$ for $\textcolor{b l u e}{b}$

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

$x = \frac{- \textcolor{b l u e}{\left(- 7\right)} \pm \sqrt{{\textcolor{b l u e}{\left(- 7\right)}}^{2} - \left(4 \cdot \textcolor{red}{8} \cdot \textcolor{g r e e n}{1}\right)}}{2 \cdot \textcolor{red}{8}}$

$x = \frac{\textcolor{b l u e}{7} \pm \sqrt{\textcolor{b l u e}{49} - 32}}{16}$

$x = \frac{7 \pm \sqrt{17}}{16}$