# How do you solve #2x^2-3x-9=0#?

##### 1 Answer

#### Answer:

#### Explanation:

Given:

#2x^2-3x-9=0#

Note that this quadratic is in the standard form:

#ax^2+bx+c=0#

with

Its discriminant

#Delta = b^2-4ac = (color(blue)(-3))^2-4(color(blue)(2))(color(blue)(-9)) = 9+72 = 81 = 9^2#

Since this is positive and a perfect square, we can deduce that this quadratic has rational zeros that we can find by factoring with integer coefficients...

Use an AC method:

Look for a pair of factors of

The pair

Use this pair to split the middle term and factor by grouping:

#0 = 2x^2-3x-9#

#color(white)(0) = (2x^2-6x)+(3x-9)#

#color(white)(0) = 2x(x-3)+3(x-3)#

#color(white)(0) = (2x+3)(x-3)#

Hence zeros:

#x=-3/2" "# or#" "x=3#