Answer
x = 30cm
Work Step by Step
Distance between $\ q_{1}$ and $\ q_{2}$ = 70cm - 20cm = 50 cm
We want $\ E_{net}$=0
So, $\ E_{1}$ = $\ -E_{2}$
$\frac{kq_{1}}{(x-20)^{2}}$ =$\frac{-kq_{2}}{(50-x)^{2}}$
$\frac{q_{1}}{(x-20)^{2}}$ = $\frac{4q_{1}}{(50-x)^{2}}$ {since$\ q_{2} = -4q_{1}$}
$\ (50-x)^{2}$ = $\ 4(x-20)^{2}$
$\ x^{2}-100x+2500$ = $\ 4x^{2}-160x+1600$
$\ x^{2}+60x+900$ = $\ 0$
So, we get x = -30cm
