Answer
a) the solutions are $5$ and $-1$
b) the solutions are $\frac{2}{3}$ and $-3$
Work Step by Step
a) $x(x-4)=5$
$x^{2}-4x=5$
$x^{2}-4x-5=0$
$(x-5)(x+1)=0$
$x-5=0$ or $x+1=0$
$x=5$ or $x=-1$
b) $x(3x+7)=6$
$3x^{2}+7x=6$
$3x^{2}+7x-6=0$
$(3x-2)(x+3)=0$
$3x-2=0$ or $x+3=0$
$3x=2$ or $x=-3$
$x=\frac{2}{3}$ or $x=-3$
