x^2+3x-28=0?

Question

SOLVE FOR X!!!! PLEASE HELP

Answer 1

1st you must factor
x^2+3x-28
(x+7)(x-4)
you can check to see if you are right by using FOIL
First x*x Outer x*-4 Inner 7*x Last 7*-4
and you get x^2-4x+7x-28
combine like terms so you get x^2+3x-28
and you know you are right b/c you ended up wih what you started with
then you set each of the factored sets equal to 0 and solve
x+7=0 x-4=0
x=-7 x=4

Answer 2

x^2 + 3x - 28 = 0

x^2 + 7x -4x - 28 =0

x ( x + 7 ) - 4 ( x + 7 ) = 0

x - 4 = 0 , x = 4

x + 7 = 0 , x = -7

Answer 3

This factors by inspection into (x+7)(x-4), so the roots are -7 and +4. Since the 28 is negative, you want two numbers whose difference is 3 and whose product is 28, and 4 and 7 work so all you need to do is to get the signs right. If this did not work, one can always use the quadratic formula.

Answer 4

If these quadratics get too hard to solve by inspection I prefer to solve by completing the square as I can never remember the equation method.

x^2 + 3x - 28 = 0

x^2 + 3x = 28

add half the coeficient of x, squared, to both sides.

x^2 + 3x + (3/2)^2 = 28 + (3/2)^2

(x + 3/2)^2 = 30.25

x + (3/2) = +/- (sqrt 30.25)

x = sqrt 30.25 - (3/2) = 4

or x= -(sqrt 30.25) - (3/2) = -7

Answer 5

(x+7)(x-4)
so x=-7 or 4

Answer 6

x^2+3x-28=0
(x+7)(x-4)=0
x+7=0
x=-7
x-4=0
x=4

roots are -7, 4

Answer 7

x² + 3x - 28

x² - 4x + 7x - 28

x(x - 4) + 7(x - 4)

(x + 7)(x - 4)

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Roots

x + 7 = 0

x + 7 - 7 = 0 - 7

x = - 7

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Roots

x - 4 = 0

x - 4 + 4 + 0 + 4

x = 4

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Answer 8

-7,4