$15 x^{2}-28=x$
Given:
$15 x^{2}-28=x$
$\Rightarrow 15 x^{2}-x-28=0$
On comparing it with $a x^{2}+b x+c=0$, we get:
$a=15, b=-1$ and $c=-28$
Discriminant $D$ is given by:
$D=\left(b^{2}-4 a c\right)$
$=(-1)^{2}-4 \times 15 \times(-28)$
$=1-(-1680)$
$=1+1680$
$=1681$
$=1681>0$
Hence, the roots of the equation are real.
Roots $\alpha$ and $\beta$ are given by :
$\alpha=\frac{-b+\sqrt{D}}{2 a}=\frac{-(-1)+\sqrt{1681}}{2 \times 15}=\frac{1+41}{30}=\frac{42}{30}=\frac{7}{5}$
$\beta=\frac{-b-\sqrt{D}}{2 a}=\frac{-(-1)-\sqrt{1681}}{2 \times 15}=\frac{1-41}{30}=\frac{-40}{30}=\frac{-4}{3}$
Thus, the roots of the equation are $\frac{7}{5}$ and $\frac{-4}{3}$.
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