use quadriatic formula to solve the equation (x+3)(x-1)=3?

Question

x=

please help me!!!!!!!!!!!

Answer 1

x² + 2x - 3 = 3
x² + 2x - 6 = 0
x = [ - 2 ±√(4 + 24) ] / 6
x = [ - 2 ± √(28) ] / 6
x = [ - 2 ± 2√(7) ] / 6
x = (1/3) [ - 1 ± √7 ]

Answer 2

first you have to get the three on the other side but you can't just add it the the rest of the equation so you multiply the (x+3)and(x-1)

(x+3)(x-1)= X(Power2) +2x+ -3=3

now you can add the 3

x(power 2) + 2x+-3+3 =

the 3 and the -3 make 0 so

x(power 2) +2x=

x(x+2)=

the answer ends up being 0 and -2

Answer 3

Expanding the brackets (FOIL):

x^2 + 2x - 3 = 3
x^2 + 2x - 6 = 0.

Now apply the quadratic formula with a = 1, b=2, c = -6

Answer 4

(x + 3)(x - 1) = 3
x^2 + 2x - 3 = 3
x^2 + 2x - 6 = 0
(x + 4)(x - 2) + 2 = 0

Answer: x^2 + 2x - 6 is a prime. If factored with (x + 4) and (x - 2) it will have a remainder of + 2.

Answer 5

x2-x+3x-3= 3
x2+2x-6=0
Use the quadratic formula , plug it in

x2 is a, 2x is b, and -6 is c

the answer is x= -2 + or - 2 root square of 7 (divided by) 2 .

Answer 6

multiply using foil then u get x squared +2x -3
then write equation x squared +2x-3=3
use reduction, x squared + 2x =0
then use quadratic formula
-b +/- root a squared minus 4 over 2a
then u should know the rest

Answer 7

(x+3)(x-1)=3
x^2 + 3x - x - 3 = 3
x^2 + 2x - 6 = 0

quadratic formula is
(-b +/- sqrt(b^2 - 4ac))/2a

so

x = (-2 +/- sqrt(4 - (4*-6)))/2
x = (-2 +/- sqrt(28))/2

so

x = -1 + sqrt(7) OR x = -1 - sqrt(7)

Answer 8

multiply it out. isolate variables.
x^2+2x-3-3=0
x^2+2x-6=0

apply quad. formula.
x= -b +/- sq.rt.:b^2 -4ac/2a
a=1 (1x^2)
b=2 (2x)
c=-6 (-6)

x= -2+/- sq.rt.: 2^2+4(1)(-6)/ 2(1)
x= -2 +/- sq.rt:4-24/2
x= -2 +/- sq.rt: -20/2
x= -2 +/- 2rad.5i/2
cancel out the 2's:
x= +rad.5 & -rad.5

rad.=sq. root.

Answer 9

The above answer is correct.