# How do you use the quadratic formula to solve #4x^2+16x+15=0#?

##### 1 Answer

#### Answer:

See explanation...

#### Explanation:

#4x^2+16x+15 = 0#

is in the form:

#ax^2+bx+c = 0#

with

The roots are given by the quadratic formula:

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

#color(white)(x) = (-16+-sqrt(16^2-4(4)(15)))/(2*4)#

#color(white)(x) = (-16+-sqrt(256-240))/8#

#color(white)(x) = (-16+-sqrt(16))/8#

#color(white)(x) = (-16+-4)/8#

#color(white)(x) = (-4+-1)/2#

That is:

#x = -5/2# or#x = -3/2#

**Footnote**

In this particular example, I would probably have favoured an AC method rather than the quadratic formula.

Given:

#4x^2+16x+15 = 0#

Find a pair of factors of

The pair

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

#0 = 4x^2+16x+15#

#color(white)(0) = (4x^2+10x)+(6x+15)#

#color(white)(0) = 2x(2x+5)+3(2x+5)#

#color(white)(0) = (2x+3)(2x+5)#

Hence roots

Note that the AC method is only applicable if the quadratic has rational zeros, whereas the quadratic formula will always give you the solution, even if the roots are Complex.