# Can you solve for x in ax^2 - bx + c = 0 (if a,b,c are constants)?

Feb 16, 2018

See a solution process below:

#### Explanation:

$a x \cdot x - b x + c = 0$ can be rewritten as:

$a {x}^{2} - b x + c = 0$

We can use the quadratic equation to solve this problem:

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

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

Substituting:

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

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

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

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

Or

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