How do you solve #5x ^ { 2} + 40x + 75= 0#?

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Answer 1 · 2017-05-19T02:35:28

#x = {-3, -5}#

You could simply the equation by factoring out a #5# in each term.

#5(x^2 + 8x + 15) = 0#

Dividing both sides by #5# gives

#x^2 + 8x + 15 = 0#

Factoring...

#(x + 3) * (x + 5) = 0#

So

#x = {-3, -5}#