How do you solve #(2w + 3) w = 35#?

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Answer 1 · 2018-04-23T00:50:46

#w = -5#

or

#w = 7/2#

#(2w+3)w = 35#

Expanding and moving all terms to one side

#2w^2 + 3w = 35#

#2w^2 +3w - 35 = 0#

Using cross multiply method;

#(2w-7)(w+5) = 0#

From this, we can use the null factor law where one of the products would have to equal #0# for this equation to be true, so;

#2w-7 = 0#

#2w = 7#

#w = 7/2#

or

#w+5 = 0#

#w = -5#