What does this Pre-Calculus problem mean

Find all the real numbers x for which x^2=10-4x

I'm going to fail my exam tomorrow :(

you just need to find x

put it into quadratic form ax^2 + bx + c = 0 then solve the quadratic

x^2 + 4x - 10 = 0

if you can't easily factor this into (x + )(x + ) then you have to use the quadratic formula

you know, the (-b +- (sqrt b^2 -4ac))/2a ?

i'll leave that up to you so you can study! look up the quadratic formula

also, whenever you are given a problem with x^2 in it and you need to find x, always rearrange it into the standard quadratic form ax^2 + bx + c = 0. once you have that, you can factor or use the quadratic formula...

good luck!

don't jump into the exam thinking you will fail. you know more than you think you know.

Set it equal to zero and then use the quadratic formula


-b +- √b^2-4(a)(c)
-------------------------------- <-- Quadratic Formula
2(a)


Answer: -2+√14 and -2-√14

It means solve for x, but the solutions must be real in order to be listed.

x² = 10 - 4x
x² + 4x - 10 = 0
Use the quadratic equation.

You'll get x = -2 ± √14

x^2 +4x - 10 = 0

x = [-4 + or - root(16 +40)]/2

x =[ -4 + or - 7.48]/2

x = -11.48/2 or 3.48/2
x = -5.74 or +1.74

solve for x:
x^2 + 4x = 10
x^2 + 4x + 4 = 10 + 4
(x + 2)^2 = 14
x + 2 = ±√14
x = -2 ±√14