These quadratic equation is killing me help me solve 2xsquared + (3 – x) 2 = 33

Use ^ to denote powers: 2xsquared = 2x^2

Also, is (3 - x) 2 supposed to be (3 - x)^2? I'll assume so.

2x² + (3 - x)² = 33
2x² + x² - 6x + 9 - 33 = 0
3x² - 6x - 24 = 0
3 (x² - 2x - 8) = 0
3 (x - 4) (x + 2) = 0

x = 4, x = -2

Mαthmφm

Um... I haven't done Algebra in probably like 8 years... but... I think it's...

2x^2 + (3 – x)2 = 33

(multiplied out): 2x^2 + 6 – 2x = 33

rewritten… 2x^2 – 2x + 6 = 33

factor out 2 from the left side to make x^2
x^2 – x + 3 = 33
subtract 33 from each side of the equation to get x^2 – x + 30 = 0

and then I kind of lose you only because I know that the two numbers to multiply to get 30 are 5 and 6… and that if you subtract 6 from 5, it will equal -1 (which is the “– x” in this equation)… so the solution is

(x – 5)(x + 6)… if you FOIL it out, you will get x^2 – x + 30 = 0

so x = 5, -6 (because you are trying to "ZERO OUT" the equation

check your answers by plugging in each of those numbers to x^2 – x + 30 = 0

i.e. for 5 .... 5^2 - 5 + 30 = 0 ? YES... because 25 - 5 + 30 = 0! yay.

AND… I could be totally wrong. Haha… Sorry. If not, check out the link below.

2X^2 + (3 - X)*2 = 33
2X^2 + 6 - 2X = 33
2X^2 - 2X - 27 = 0
then using the formula
2+/-the root of -2^2 - 4*-2*-27 all divided by 4
2+/- the root of -212 over 4
so the answers are 0.5 + 3.64i and 0.5 - 3.64i

x = 1/2 (1 - Sqrt[55]) || x = 1/2 (1 + Sqrt[55])