Answer
$x=3$ and $x=-\dfrac{1}{5}$
Work Step by Step
$5x^{2}-3=14x$
Take the $14x$ to the left side of the equation:
$5x^{2}-14x-3=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For this equation, $a=5$, $b=-14$ and $c=-3$
Substitute:
$x=\dfrac{-(-14)\pm\sqrt{(-14)^{2}-4(5)(-3)}}{2(5)}=\dfrac{14\pm\sqrt{196+60}}{10}=...$
$...=\dfrac{14\pm\sqrt{256}}{10}=\dfrac{14\pm16}{10}$
Our two solutions are:
$x=\dfrac{14+16}{10}=3$
$x=\dfrac{14-16}{10}=-\dfrac{1}{5}$
