How do you solve #9x^2 + 6x + 1= 0# using the quadratic formula?

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Answer 1 · 2017-06-06T21:07:20

See a solution process below:

For #a^x2 + bx + c = 0#, the values of #x# which are the solutions to the equation are given by:

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

Substituting #9# for #a#; #6# for #b# and #1# for #c# gives:

#x = (-6 +- sqrt(6^2 - (4 * 9 * 1)))/(2* 9)#

#x = (-6 +- sqrt(36 - 36))/(2* 9)#

#x = (-6 +- sqrt(0))/(2* 9)#

#x = (-6 +- 0)/(2* 9)#

#x = -6/18#

#x = -1/3#