Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - 10.3 - Quadratic Formula - Problem Set 10.3: 57

Answer

{$-3.55,1.22$}

Work Step by Step

Step 1: Comparing $3x^{2}+7x-13=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=3$, $b=7$, and $c=-13$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a,b, and c in the formula, we obtain: $x=\frac{-(7) \pm \sqrt {(7)^{2}-4(3)(-13)}}{2(3)}$ Step 4: $x=\frac{-7 \pm \sqrt {49+156}}{6}$ Step 5: $x=\frac{-7 \pm \sqrt {205}}{6}$ Step 6: $x=\frac{-7 \pm 14.318}{6}$ Step 7: $x=\frac{-7+14.318}{6}$ or $x=\frac{-7-14.318}{6}$ Step 8: $x=\frac{7.318}{6}$ or $x=\frac{-21.318}{6}$ Step 9: $x=1.220\approx1.22$ or $x=-3.553\approx-3.55$ Step 10: Therefore, the solution set is {$-3.55,1.22$}.
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