Question

3x²+10x-8=0 by completing square method
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Answer 1

[tex]since you have a number greater than 1 before the x2 you will need to multiply that by the −8 at the end to get true divisible numbers. 3 x -8 = -24, ergo you need to find two numbers that multiply to get -24 and add up to reach -10 (your middle term). in this case, it would be 2 and -12.

align the equation to make further factoring easier:

3x2−10x−8=0

3x2−12x+2x−8=0 note that (3x2+2x−12x−8=0) isn't as user friendly

factor the (3x2−12x+2x−8=0) into two sets of brakcets:

(3x(x−4)+2(x−4)=0) and group them together:

(3x+2)(x−4)=0

now you need to determine what values for x will make this equation true.