How do you solve #12u ^ { 2} - 5u - 3= 0#?

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Answer 1 · 2017-05-02T20:04:57

#u = 3/4 or u =-1/3#

There are three ways to solve a quadratic equation in Algebra:

Let's use factorising in this example.

#12u^2 -5u -3=0#

Find factors of #12 and 3# whose products subtract to give #5#

#(4u -3)(3u +1)=0#

Either factor could be equal to #0#

If #4u -3 =0," "rarr u= 3/4#

If #3u+1=0" "rarr u = -1/3#