How do you solve # 4x^2 + 23x + 15 = 0#?

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Answer 1 · 2016-05-08T22:25:41

Factorise, then let each factor in turn be equal to 0.

#x = -3/4 or x = -5#

#4x^2 + 23x +15 = 0 " "# Find factors of 4 and 15 which add to 23

#(4x + 3)(x + 5) = 0" "# 4x5 + 1x3 = 23

if # 4x + 3 = 0 rArr x = -3/4#
if # x + 5 = 0 rArr x = -5#

Answer 2 · 2016-05-08T22:27:14

#x = -5# or #x=-3/4#

Use the quadratic formula.

The equation:

#4x^2+23x+15 = 0#

is in the form #ax^2+bx+c = 0#, with #a=4#, #b=23# and #c=15#.

This has roots given by the quadratic formula:

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

#=(-23+-sqrt(23^2-(4*4*15)))/(2*4)#

#=(-23+-sqrt(529-240))/8#

#=(-23+-sqrt(289))/8#

#=(-23+-17)/8#

That is:

#x = (-40)/8 = -5#

or:

#x = (-6)/8 = -3/4#