  ### Home > CCAA8 > Chapter 9 Unit 10 > Lesson CCA: 9.1.3 > Problem9-28

9-28.

Use the Quadratic Formula to solve part (b) of problem 9-27 above. Did your solution match the solution you got by factoring and using the Zero Product Property (in part (b) of problem 9-27)?

Example 2: Solve $3x^2+x-14=0$ for $x$ using the Quadratic Formula.

Solution: This method works for any quadratic.  First, identify $a,b$, and $c$$a$ equals the number of $x^2$-terms, $b$ equals the number of $x$-terms, and $c$ equals the constant. For $3x^2+x-14=0$, $a=3$, $b=1$, and $c=-14$.  Substitute the values of $a,b$, and $c$ into the Quadratic Formula and evaluate the expression twice: once with addition and once with subtraction. Examine this method below:

$\begin{array}{l} x = \frac{-1 + \sqrt{1^2 - 4 \left(3\right) \left(-14\right)}}{2 \cdot 3} \\ \phantom{x} = \frac{-1 - \sqrt{169}}{6} \\ \phantom{x} = \frac{12}{6} \\ \phantom{x} = 2 \end{array} \quad \text{or} \quad \begin{array}{l} x = \frac{-1 + \sqrt{1^2 - \left(3\right) \left(-14\right)}}{2 \cdot 3} \\ \phantom{x} = \frac{-1 - \sqrt{169}}{6} \\ \phantom{x} = \frac{-14}{6} \\ \phantom{x} = -\frac{7}{3} \end{array}$