How do you solve #x^2 - 3x = -3# using the quadratic formula?

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Answer 1 · 2017-05-31T16:15:08

#x=(3+- sqrt(-3))/2=3/2pm1/2isqrt3#

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

the quadratic equation needs to be in the following form:

#ax^2 + bx + c = 0#

where a, b and c is any number.

Finally, we plug the values in the formula and solve for #x#.

Step 1: Get the equation in the right format:

#x^2-3x+3=0#

#a=1#
#b=-3#
#c=3#

Step 2: Plug the values into the quadratic formula:

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

#x=(3+- sqrt(9-12))/2#

#x=(3+- sqrt(-3))/2=3/2pm1/2isqrt3#

There are no real roots to this quadratic - instead there are two imaginary roots.