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How do you solve #x( x + 3) - 40= 0#?

1 Answer

Apr 16, 2018

x=5, x=-8

Explanation:

First expand brackets:

#x(x+3)-40=0# becomes #x^2+3x-40=0#

Then solve as normal quadratic:

#x^2+3x-40=0#
#(x-5)(x+8)=0#

If #x-5=0# then #x=5#
If #x+8=0# then #x=-8#

Hope this helps!

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