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7-28.

Use the Quadratic Formula to solve part (b) of problem 7-27. Did your solution match the solution you got in part (b) of problem 7-27? Which solution method do you prefer?

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,\ \text{and } c$$a$ equals the number of $x^2$ terms, $b$ equals the number of $x$ terms, and $c$ equals the constant. For $3x^2+x-14=0$$1=3,\ b=1,\ \text{and } c=-14$. Substitute the values of $a,b,\text{and } c$ into the Quadratic Formula and evaluate the expression twice: once with addition and once with subtraction. Examine this method below:

$x=\frac{-1+\sqrt{1^2-4(3)(-14)}}{2\cdot3} \\ =\frac{-1+\sqrt{169}}{6} \\ =\frac{12}{6}=2$     or      $x=\frac{-1-\sqrt{1^2-4(3)(-14)}}{2\cdot3} \\ =\frac{-1-\sqrt{169}}{6} \\ =\frac{-14}{6}=-\frac{7}{3}$