### Home > CCA > Chapter 9 > Lesson 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:

$\begin{array}{l c l} {x=\frac{-1+\sqrt{1^2-4(3)(-14)}} {2⋅3}\\=\frac{-1-\sqrt{169}}{6}\\=\frac{12}{6}=2} &\text{ or }&{x=\frac{-1+\sqrt{1^2-4(3)(-14)}} {2⋅3}\\ =\frac{-1-\sqrt{169}}{6}\\=\frac{-14}{6}=-\frac{7}{3}}&&\\& \end{array}$