Chapter 5 Polynomials and Polynomial Functions - 5.5 Apply the Remainder and Factor Theorems - 5.5 Exercises - Mixed Review - Page 368: 55

$x=\frac{-15 \pm i\sqrt 23}{4}$

Given: $2x^2+15x+31=0$ The equation is in quadratic form $ax^2+bx+c=0$ with $a=2, b=15, c=31$. Hence, the solution of the equation is: $x=\frac{-b\pm\sqrt b^2-4ac}{2a}$ $x=\frac{-15 \pm \sqrt 15^2-4.2.31}{2.2}$ $x=\frac{-15 \pm \sqrt -23}{4}$ $x=\frac{-15 \pm i\sqrt 23}{4}$