Answer
$\left\{ -\dfrac{1}{5},3 \right\}$
Work Step by Step
The standard form of the given equation, $
5x^2-3=14x
,$ is
\begin{array}{l}\require{cancel}
5x^2-14x-3=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the given quadratic equation, $
5x^2-14x-3=0
,$ are
\begin{array}{l}\require{cancel}
\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\pm\sqrt{(16)^2}}{10}
\\\\=
\dfrac{14\pm16}{10}
\\\\=
\dfrac{14-16}{10}
\text{ OR }
\dfrac{14+16}{10}
\\\\=
\dfrac{-2}{10}
\text{ OR }
\dfrac{30}{10}
\\\\=
-\dfrac{1}{5}
\text{ OR }
3
.\end{array}
Hence, the solutions are $
\left\{ -\dfrac{1}{5},3 \right\}
.$
