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: 56

Answer

{$-2.77,1.27$}

Work Step by Step

Step 1: Comparing $2x^{2}+3x-7=0$ to the standard form of a quadratic equation, $ax^{2}+bx+c=0$, we obtain: $a=2$, $b=3$, and $c=-7$ 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: $x=\frac{-(3) \pm \sqrt {(3)^{2}-4(2)(-7)}}{2(2)}$ Step 4: $x=\frac{-3 \pm \sqrt {9+56}}{4}$ Step 5: $x=\frac{-3 \pm \sqrt {65}}{4}$ Step 6: $x=\frac{-3 \pm 8.062}{4}$ Step 7: $x=\frac{-3+8.062}{4}$ or $x=\frac{-3-8.062}{4}$ Step 8: $x=\frac{5.062}{4}$ or $x=\frac{-11.062}{4}$ Step 9: $x=1.266\approx1.27$ or $x=-2.766\approx-2.77$ Step 10: Therefore, the solution set is {$-2.77,1.27$}.
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