How do you use the quadratic formula to find both solutions to the quadratic equation #3x^2 + 6x = 12#?

1 Answer
Apr 27, 2018

Answer:

#x = -1 +- sqrt5#

Here's how I did it:

Explanation:

#3x^2 + 6x = 12#

To use the quadratic formula, we first have to set the quadratic equal to zero.

To do this, let's subtract #color(red)(12)# from both sides of the equation:
#3x^2 + 6x quadcolor(red)(-quad12) = 12 quadcolor(red)(-quad12)#

#3x^2 + 6x - 12 = 0#

Now, the equation is in standard form, or #color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c)#, so:
#color(red)(a = 3)#, #color(magenta)(b = 6)#, and #color(blue)(c = -12)#

The quadratic formula can solve for #x# in any case, and we use it especially when the expression is not factorable.
The quadratic formula is:
#x = (color(magenta)(-b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a))#

Now we can plug in our values for #a#, #b#, and #c#:
#x = (color(magenta)(-6) +- sqrt((color(magenta)(-6))^2 - 4(color(red)(3))(color(blue)(-12))))/(2(color(red)(3)))#

Simplify:
#x = (-6 +- sqrt(36 - 4(-36)))/6#

#x = (-6 +- sqrt(36 + 144))/6#

#x = (-6 +- sqrt(180))/6#

#x = (-6 +- 6sqrt(5))/6#

Divide by the denominator to get the simplified answer:
#x = -1 +- sqrt5#

Hope this helps!