Answer
$x=\dfrac{1}{5} ,-3 $
Work Step by Step
Put all terms on the left side:
$5 x^{2}+2=5-14 x$
$5 x^{2}+2-5+14 x=0$
$5 x^{2}+14 x-3=0$
Solve using the Quadratic Formula $x=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2a}$ with $a=5, b=14, \text{ and } c=-3$:
$x=\dfrac{-14 \pm \sqrt{14^{2}-4(5)(-3)}}{2(5)}$
$x=\dfrac{-14 \pm \sqrt{196+60}}{10}$
$x=\dfrac{-14 \pm \sqrt{256}}{10}$
$x=\dfrac{-14 \pm 16}{10}$
$\begin{aligned} x&=\frac{-14+16}{10}(or) \frac{-14-16}{10} \\ x&=\frac{2}{10}\quad \text{or} \quad \frac{-30}{10} \\ x &=\frac{1}{5} \quad \text{or} \quad -3\end{aligned}$
