Can you please review this to see if I did this right?
2cos^2x+3cosx=0
sq. root of (cos^2x)+(cosx)=sq. root of (-3/2)
2cos=(sq. root of -3 times sq. root of 2)/2)) times 2/1
cos=(2 times sq. root of -3 times sq. root of two)/2))
2cos^2x+3cosx=0
sq. root of (cos^2x)+(cosx)=sq. root of (-3/2)
2cos=(sq. root of -3 times sq. root of 2)/2)) times 2/1
cos=(2 times sq. root of -3 times sq. root of two)/2))