algebra Q's?

Question

how do u go about solving:
x^2 + 21x= 22


(x^ is x squared)

Answer 1

x^2 + 21x= 22
First subtract 22 from each side
x^2 + 21x - 22 = 0
What factors of -22 add up to 21?
22 - 1 = 21 and 22 times -1 is -22
(x+22)(x-1) = 0
x+22 = 0 or x-1 = 0
x = -22 or x = 1

Answer 2

x^2 + 21x = 22
x^2 + 21x - 22 = 0
Now factorise the polynomial
x^2 + 21x - 22 = 0
x^2 - x + 22x - 22 = 0
x(x - 1) + 22(x - 1) = 0
(x + 22)(x - 1) = 0
Now either x + 22 = 0
or x - 1 = 0
Put x + 22 = 0
x = -22
Put x - 1 = 0
x = 1
Two possible values of 'x' are 1 and -22

Answer 3

Good Evening,

x^2 + 21x = 22 Given
x^2 + 21x - 22 = 0 (addition of equalities)
(x+22)(x-1) = 0 (Factor - choose your favorite method)
x = -22 or x=1 (by zero property)
x = {-22, 1}

~i~

Answer 4

Move the 22 to the other side then factor the equation into (x+22) and (x-1) then solve those two equations for x.

Answer 5

Bring the 22 over to the left-hand side, and use the quadratic formula.

Answer 6

x^2+21x-22=0
x^2+22x-x-22=0
x(x+22)-1(x+22)=0
(x-1)(x+22)=0
x=1 and
x=-22

Answer 7

using the quadratic formula:
x=[ -21 +- sqrt ( 21^2 -4 (-22) ) ]/2
= [-21 + - 23 ]/2
x= [-21 +23]/2 or x= [-21 -23 ]/2
x = 1, x= -22
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Answer 8

You have to use the quadratic formula search it up. its

x = -b +/- √ b² - 4ac
______________
2a

you will end up with

x = 1 and x = -22

there are always 2 answers in a quadratic equation