Answer
$x = \frac{-1 ± \sqrt {61}}{10}$
Work Step by Step
Let's rewrite this equation so that all terms are on the left side and $0$ is on the right side:
$5x^2 + x - 3 = 0$
We are asked to solve this equation using the quadratic formula, which is given by: $x = \frac{-b ± \sqrt {b^2 - 4ac}}{2a}$ where $a$ is the coefficient of the first term, $b$ is the coefficient of the 1st degree term, and $c$ is the constant. In this exercise, $a = 5$, $b = 1$, and $c = -3$.
Let's plug in the values for $a$, $b$, and $c$ from our equation into the quadratic formula:
$x = \frac{-(1) ± \sqrt {(1)^2 - 4(5)(-3)}}{2(5)}$
Let's simplify:
$x = \frac{-1 ± \sqrt {1 + 60}}{10}$
Let's simplify what is inside the radical:
$x = \frac{-1 ± \sqrt {61}}{10}$
