Check the picture below.
[tex]\textit{area of a triangle}\\ A = \cfrac{1}{2}bh~~ \begin{cases} A = area\\ b = base\\ h = height\\[-0.5em] \hrulefill\\ A = 7\\ h = 2b - 3 \end{cases}\implies 7 = \cfrac{1}{2}b(2b-3)\implies 7=\cfrac{b(2b-3)}{2} \\\\\\ 14 = b(2b-3)\implies 14 = 2b^2-3b\implies 0 = 2b^2-3b-14[/tex]
[tex]0 = (b+2)(2b-7)\implies\blacktriangleright b= \begin{cases} -2\\ \frac{7}{2}~~\textit{\large \checkmark} \end{cases}\blacktriangleleft \\\\\\ \stackrel{\textit{we know that}}{h = 2b - 3}\implies h = 2\left( \frac{7}{2} \right)-3\implies h = 7 - 3\implies \blacktriangleright h = 4 \blacktriangleleft[/tex]
