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:

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

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