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

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Answer 1 · 2018-04-16T08:44:10

x=5, x=-8

First expand brackets:

$x (x + 3) - 40 = 0$ $x(x+3)-40=0$ becomes $x^{2} + 3 x - 40 = 0$ $x^2+3x-40=0$

Then solve as normal quadratic:

$x^{2} + 3 x - 40 = 0$ $x^2+3x-40=0$
$(x - 5) (x + 8) = 0$ $(x-5)(x+8)=0$

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

Hope this helps!

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