How do you find the roots, real and imaginary, of #y= 12x^2 + 35x + 72 # using the quadratic formula?

1 Answer
Feb 13, 2018

Answer:

#x = -35/24+-sqrt(2231)/24i#

Explanation:

The equation:

#y = 12x^2+35x+72#

is a quadratic in standard form:

#y = ax^2+bx+c#

with #a=12#, #b=35# and #c=72#

It has zeros given by the quadratic formula:

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

#color(white)(x) = (-(color(blue)(35))+-sqrt((color(blue)(35))^2-4(color(blue)(12))(color(blue)(72))))/(2(color(blue)(12)))#

#color(white)(x) = (-35+-sqrt(1225-3456))/24#

#color(white)(x) = (-35+-sqrt(-2231))/24#

#color(white)(x) = -35/24+-sqrt(2231)/24i#

Note that #2231 = 23 * 97# has no square factors, so #sqrt(2231)# cannot be simplified.