A point charge q1 = -4.00 nC is at the point x = 0.60 m, y = 0.80 m , and a second point charge q2 = +6.00 nC?

the 'pros' probably wouldn't like this approach

but

since field strength is N/C

place an imaginary 1C charge at the origin and calculate the force of each of the given charges on the charge at the origin

then add them (they are vectors) to get the reluctant which is the field strength at the origin

use F = K q1 12 / r^2

another trick is to resolve any pushes into pulls

let our imaginary charge be +1C, then -4nC exerts a pull but +6nC pushes so reflect it's position across the origin to (-0.60 , 0) and change the sign of the charge to negative

(0.60 , 0.80) makes a 3,4,5 rt. triangle with hypotenuse of 1.0 m

and tan(angle) = 0.80 /0.60 = 1.333, angle - 53.1° above the +x-axis

F = 8.99E9 (4.00E-9) (1) / 1.00^2 = 35.96 N @ 53.1° (first force)

then

F = 8.99E9 (6.0E-9) (1) / 0.60^2 = 149.8 N @ 180° (second force)

add the two forces by constructing a parallelogram with the two vectors as adjacent sides

then find (r, theta) using Law of Cosines and Law of Sines

I get

35.96 N @ 53.1 + 149.8 N @ 180° = 131 N @ 167.4°

which translates to

field strength = 131 N/C at an angle of 12.6° above the negative y-axis