$$(6-2) \times [-5+2-8/4-3 \times (2-3-6/2)]$$
$28$
$$4\left(-5+2-\frac{8}{4}-3\left(2-3-\frac{6}{2}\right)\right)$$
$$4\left(-3-\frac{8}{4}-3\left(2-3-\frac{6}{2}\right)\right)$$
$$4\left(-3-2-3\left(2-3-\frac{6}{2}\right)\right)$$
$$4\left(-5-3\left(2-3-\frac{6}{2}\right)\right)$$
$$4\left(-5-3\left(-1-\frac{6}{2}\right)\right)$$
$$4\left(-5-3\left(-1-3\right)\right)$$
$$4\left(-5-3\left(-4\right)\right)$$
$$4\left(-5-\left(-12\right)\right)$$
$$4\left(-5+12\right)$$
$$4\times 7$$
$$28$$
Show Solution
Hide Solution
$2^{2}\times 7$
