Daiana L.

asked • 03/06/19

How do I solve x² + 2x − 8 = 0?

Eric W.

(x+4)*(x-2)=0, then X=-4 AND X=2.



when the variable is to the 2nd power you have a quadratic equation. The standard form ax^2 + bx + c where a is not equal to zero. to solve using the formula : x = -b + or - the square root of (b^2 - 4*a*c) all over 2*a here you have a=1, b=2 and c = -8 substitute the values in the formula and simplify. you should get the same answer as above. When a = 1 it is faster to solve by factoring but the formula works for any quadratic.


Vivek A.

The equation of finding X= Square root of (b^2 +or- 4*a*c), Here b=2, a=a and c=-8. And also you can use (x+4)*(x-2)=0,


San P.

Ok I'll respond way simpler than any of the experts you hear. They are right, just not easy. Theres a really easy way to know this. x^2 is always going to be A x is always going to be B And any plain number will always be C So just plug it into the Quadratic formula { (-b±√(b^2-4ac))/2a } The ± means plus or minus so you have to do this equation 2x one with + and one with -. Solve it and you get {-4, 2} in Set Builder Notation as x is equal to 2 things. None of these Experts know it the easy way lol.


Mohammad U.

There are more possible ways to do it, without using quadratic formula I will do it in a simplest way, multiply X^2 with -8 in a rough work(RW), which will be -8x^2, now find the possible ways by multiplying it should give -8x^2, and by adding or subtracting it give 2x, and like 1*8, 2*4, look there are more possible ways, but have to pick that one which satisfies our requirement, we will pick 2*4. The equation will become like this: (x^2)+4x-2x-8=0 now check whether it is satisfying the conditions or not, like by the terms in the middle by multiplying it is giving -8x^2, and by adding +2x, yes it is satisfying the terms. now taking commons from right 2 and left 2 terms, x(x+4)-2(x+4)=0 taking x+4 common, the equation will become: (x+4)(x-2)=0 now taking x+4=0 and x-2=0 so it becomes: x= -4 AND x=2 S.S= {-4,2}


2 Answers By Expert Tutors


Eric H. answered • 02/12/21

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