How do you solve #2a^2 - 9a = 5#?

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Answer 1 · 2016-03-22T20:15:29

#a=-1/2" or "a=-5#

Given:#" "2a^2-9a=5#

Write as:#" "2(a^2-9/2a-5/2)=0#

Divide both sides by 2

#a^2-9/2a-5/2=0#

As there is no coefficient in front of #a^2# there are two possible starting point

#(-a+?)(-a+?)" or "(a+?)(a+?)#

Lets investigate positive #a#

Notice that there are halves in this equation and we have a constant of #5/2#

Lets test #1/2xx5# giving

#(a+1/2)(a-5)-> a^2+1/2 a -5a-5/2#

But #1/2 a-5a -> 1/2 a-10/2 a= -9/2 a" "# Which is what we need.
#color(blue)((a+1/2)(a-5)-> a=-1/2" or "a=-5)#