$4\cdot(x+\dfrac{3}{7})\cdot (x-\dfrac{1}{8})=0$

$\dfrac{1}{8}x^2=18$

$5x^2=-4x$

$-12x+32=-x^2$

$\dfrac{5}{2}x^2+\dfrac{7}{8}x-\dfrac{3}{8}=0$

$3 \cdot\left(x-\dfrac{1}{7}\right)^2-12=0$

$2x^2-72 =0$

$-6x^2=\dfrac{1}{3}x$

$(8x+12)\cdot(3x-15)=0$

$-2 \cdot\left(x+\dfrac{1}{2}\right)^2+4=0$