Нужно решить уравнение 2x^2-3x-20/6x^2-20x-16=(6x-4)^2/36x^2-16

$\frac{2x^2-3x-20}{6x^2-20x-16} =\frac{(6x-4)^2}{36x^2-16} \\\frac{2x^2-3x-20}{2(3x^2-10x-8)} =\frac{(6x-4)^2}{(6x-4)(6x+4)}$

2x² - 3x - 20 = 0
D = 9 + 160 = 169
$x_1= \frac{3-13}{4}= -\frac{5}{2} \\ x_2= \frac{3+13}{4}=4$
2x² - 3x - 20 = 2·(x + 5/2)·(x - 4)

3x² - 10x - 8 = 0
D = 100 + 96 = 196
$x_1= \frac{10-14}{6}= -\frac{2}{3} \\ x_2= \frac{10+14}{6}=4$
3x² - 10x - 8 = 3·(x + 2/3)·(x - 4)

$\frac{2(x+ \frac{5}{2})(x-4)}{2*3(x+ \frac{2}{3})(x-4)} =\frac{(6x-4)^2}{(6x-4)(6x+4)}\\ \frac{(2x+5)(x-4)}{(6x+4)(x-4)} =\frac{(6x-4)^2}{(6x-4)(6x+4)} \\ \\ \frac{2x+5}{6x+4} =\frac{6x-4}{6x+4} \\ x-4 \neq 0 \\ 6x-4 \neq 0 \\ \\2x+5=6x-4\\ x-4 \neq 0 \\ 6x-4 \neq 0 \\ 6x+4 \neq 0 \\ \\4x=9\\ x \neq 4 \\ 6x \neq 4 \\ 6x \neq -4 \\ \\x= \frac{9}{4} \\ x \neq 4 \\ x \neq \frac{2}{3} \\ x \neq -\frac{2}{3} \\ \\ x= \frac{9}{4}$