Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 71



Work Step by Step

$3x^{2}+6x-5=0$ Use the quadratic formula to solve this equation. This formula is: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ For this particular equation, $a=3$, $b=6$ and $c=-5$ $x=\dfrac{-6\pm\sqrt{(6)^{2}-4(3)(-5)}}{2(3)}=\dfrac{-6\pm\sqrt{36+60}}{6}=\dfrac{-6\pm\sqrt{96}}{6}$ $x=\dfrac{-6\pm4\sqrt{6}}{6}=-1\pm\dfrac{2}{3}\sqrt{6}$
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