## Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole

# Chapter 0 - Section 0.5 - Solving Polynomial Equations - Exercises - Page 30: 20

#### Answer

Solution set: $\displaystyle \{-1, \frac{3}{2}\}$

#### Work Step by Step

Quadratic formula, a=$2$, b=$-1$, c=$-3$ $x=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ Checking $\Delta=b^{2}-4ac=1-4(2)(-3)=1+24=25$ $\Delta$ is positive, so there will be 2 real solutions. $x=\displaystyle \frac{-(-1)\pm\sqrt{25}}{2(2)}=\frac{1\pm 5}{4}$ $x=\displaystyle \frac{1-5}{4}=\frac{-4}{4}=-1$ or $x=\displaystyle \frac{1+5}{4}=\frac{6}{4}=\frac{3}{2}$ Solution set: $\displaystyle \{-1, \frac{3}{2}\}$

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