SSC CHSL 19 Apr 2021 Shift 3 Paper

Let the length of the base of the triangle be $\mathit{x}\mathit{c}\mathit{m}$.
$\therefore$ According to the question,
$\Rightarrow \frac{1}{2}\times$ Base $\times$ Altitude (Height)
$=$ Area of triangle
$\Rightarrow \frac{1}{2}\times \mathit{x}\times \left(\mathit{x}+3\right)=104$
$\Rightarrow {\mathit{x}}^2+3\mathit{x}=208$
$\Rightarrow {\mathit{x}}^2+3\mathit{x}-208=0$
$\Rightarrow {\mathit{x}}^2+16\mathit{x}-13\mathit{x}-208=0$
$\Rightarrow \mathit{x}\left(\mathit{x}+16\right)-13\left(\mathit{x}+16\right)=0$

Since, the length can not be negative,