How do you solve #-15x^2 - 10x + 40 = 0 # using the quadratic formula?

1 Answer
Aug 14, 2015

Answer:

#x_(1,2) = -(1 +-5)/3#

Explanation:

For a general form quadratic equation

#color(blue)(ax^2 + bx + c = 0)#

the two possible roots of the equation take the form - this is the quadratic formula

#color(blue)(x_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)#

Now, your quadratic equation looks like this

#-15x^2 - 10x + 40 = 0#

You can divide all the terms by #5# to simplify it

#-3x^2 - 2x + 8 = 0#

In your case, you have #a=-3#, #b=-2#, and #c=8#. THis means that the two solutions will be

#x_(1,2) = (-(-2) +- sqrt( (-2)^2 - 4 * (-3) * 8))/(2 * (-3))#

#x_(1,2) = (2 +- sqrt(100))/(-6)#

#x_(1,2) = (2 +- 10)/(-6) = -(1 +- 5)/3 = {(x_1 = (-1 - 5)/3 = -2), (x_2 = (-1 + 5)/3 = 4/3) :}#

Your quadratic equation will thus have two roots, #x_1 = color(green)(-2)# and #x_2 = color(green)(4/3)#.