Question

using the quadratic formula to solve $5x = 6x^2 - 3$, what are the values of $x$?
$\frac{5\pm3\sqrt{11}}{12}$
$\frac{5\pm\sqrt{87}}{12}$
$\frac{5\pm\sqrt{47}}{12}$
$\frac{-5\pm\sqrt{97}}{12}$

Explanation:

Step1: Rewrite in standard form

Rewrite \(5x = 6x^2 - 3\) as \(6x^2 - 5x - 3 = 0\). Here, \(a = 6\), \(b = -5\), \(c = -3\).

Step2: Apply quadratic formula

Quadratic formula: \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 6\), \(b = -5\), \(c = -3\):
\(x=\frac{-(-5)\pm\sqrt{(-5)^2 - 4(6)(-3)}}{2(6)}=\frac{5\pm\sqrt{25 + 72}}{12}=\frac{5\pm\sqrt{97}}{12}\).

Answer:

\(\frac{5\pm\sqrt{97}}{12}\) (corresponding to the option with this expression)