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.

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