## Calculus with Applications (10th Edition)

x = $\frac{-2-\sqrt 7}{3}$, $\frac{-2+\sqrt 7}{3}$
$3x^{2}+4x=1$ First, you must move the constant to the left and change its sign: $3x^{2}+4x-1=0$ Then you must solve the quadratic equation ($ax^{2}+bx+c=0$) using $x=\frac{-b±\sqrt 4^{2}-4\times3\times-1}{2\times3}$ Simplify everything in the square root and denominator to end up with: $x=\frac{-4±\sqrt 28}{6}$ Simplify the radical: $x=\frac{-4±2\sqrt 7}{6}$ Separate the positive and negative solutions and simplify the fraction to get: $\frac{-2-\sqrt 7}{3}$ and $\frac{-2+\sqrt 7}{3}$ which both equal x