How do you solve #0=2x^2 + 5x - 12# using the quadratic formula?

1 Answer
Oct 19, 2015

Answer:

#x_(1,2) = (-5 +- 11)/4#

Explanation:

For a general form quadratic equation

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

its roots can be determined using the quadratic formula

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

In your case, you have

#2x^2 + 5x - 12 = 0#

which implies that #a = 2#, #b = 5#, and #c = -12#.

The two roots will thus take the form

#x_(1, 2) = (-5 +- sqrt(5^2 - 4 * 2 * (-12)))/(2 * 2)#

#x_(1,2) = (-5 +- sqrt(121))/4#

#x_(1,2) = (-5 +- 11)/4 = {(x_1. = (_5 - 11)/4 = -4), (x_2 = (-5 + 11)/4 = 3/2) :}#

The two solutions to this quadratic equation will thus be

#x_1 = color(green)(-16)" "# and #" "x_2 = color(green)(3/2)#