How do you find the zeros, real and imaginary, of # y=x^2/3+1/7x-5/9 # using the quadratic formula?

#x=(-3+-sqrt(1007/3))/14#

Explanation:

The quadratic formula states that the zeros of a quadratic function in the form #y=ax^2+bx+c# are equal to

#x=(-b+-sqrt(b^2-4ac))/(2a)#

With the quadratic #y=x^2/3+1/7x-5/9#, we can see that

#{(a=1/3),(b=1/7),(c=-5/9):}#

Plug these into the quadratic formula:

#x=(-1/7+-sqrt((1/7)^2-(4xx1/3xx-5/9)))/(2xx1/3)#

Simplify.

#x=(-1/7+-sqrt(1/49+20/27))/(2/3)#

#x=(-1/7+-sqrt(27/1323+980/1323))/(2/3)#

#x=(-1/7+-sqrt(1007/1323))/(2/3)#

#x=(-1/7+-sqrt(1007/(9xx49xx3)))/(2/3)#

#x=(-1/7+-1/21sqrt(1007/3))/(2/3)#

#x=(-1/7+-1/21sqrt(1007/3))*(3/2)#

#x=(-3/7+-1/7sqrt(1007/3))/2#

#x=(-3+-sqrt(1007/3))/14#