Algebra Help?

Question

Can someone help me solve this and explain it to me in the process please.

Solve 2x^2-3x-2=0 using the quadratic formula.

Answer 1

[-b +/- sqrt ( b^2 - 4ac) ] /2a
general form

[ 3 +/- sqrt( 3^2 - 4(2)(-2) ) ] / 2(2)
plug in values for a,b,c

[3 +/- sqrt(9 + 16)] /4
simplify

[3 +/- sqrt(25)]/ 4
simplify

[3 +/- 5] /4
simplify

-2/4 or 8/4
solve

x = 2 or -1/2

Answer 2

Given: 2x^2 - 3x - 2 = 0

Answer: (x-2) (2x+1)

proof: (x-2) (2x+1)
*multiply each.. do the long method! therefore,
2x^2 + x - 4x - 2 = 0; simplify
2x^2 - 3x - 2= 0

Answer 3

*Quadratic Formula: [- b +/- √(b^2 - 4ac)] / 2a

First: substitute the numbers with the corresponding variables...a = 2, b = -3 & c = -2

[- (-3) +/- √((-3)^2 - 4(2)(-2))] / 2(2)
[3 +/- √((-3)(-3) - 4(-4))] / 4
[3 +/- √(9 - (-16))] / 4
[3 +/- √(9+16)] / 4
[3 +/- √(25)] / 4
[3 +/- 5] / 4

Sec: you will always have 2 solutions.

a. [3 + 5] / 4 = 8/4 = 2
b. [3 - 5] / 4 = -2/4 = -1/2
Solutions: 2 & -1/2

Answer 4

2x^2-3x-2=0
2x^2-4x+x-2=0
2x(x-2)+1(x-2)=0
(x-2)(2x+1)=0
The product of two numbers or expression cannot be zero,unless one of them is zero
Therefore either x-2=0 or 2x+1=0
If x-2=0,then x=2
If 2x+1=0,then 2x= -1 or x= -1/2
x=2 or -1/2
[-b +/- sqrt ( b^2 - 4ac) ] /2a
general form

[ 3 +/- sqrt( 3^2 - 4(2)(-2) ) ] / 2(2)
plug in values for a,b,c

[3 +/- sqrt(9 + 16)] /4
simplify

[3 +/- sqrt(25)]/ 4
simplify

[3 +/- 5] /4
simplify

-2/4 or 8/4
solve

x = 2 or -1/2
ITS 1 OF UM GO TO GOOGLE AND FIGURE IT OUT!!!

Answer 5

2X^2 -3X -2 =0

Add the 2 to the 0...

2X^2 -3X =2

Square root that 2 to get the "^2" out of the way//

2X - 3X = 1.41421356

2x - 3x =1x

1X = 1.41421356

divide 1.41421356 by 1

X = 1.41421356

Answer 6

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

x - 2 = 0, x = 2
2x + 1 = 0, 2x = - 1, x = - 1/2

Answer: x = 2, - 1/2

Answer 7

Your question is -

2x^2-3x-2=0

Using the quadratic formula we get -

x = [ - ( -3 ) ( plus,minus ) Sq.root of { ( - 3 )^2 - 4 (2).(-2) } ] divided by 2(2)

x = [ 3 ( plus,minus ) Sq.root of { 9 + 16 } ] divided by 4

x = [ 3 (plus , minus ) 5 } ] divided by 4
ie x = 2 or ( - 1/2 ) .................................... Answer

Answer 8

[06]
2x^2-3x-2=0
2x^2-4x+x-2=0
2x(x-2)+1(x-2)=0
(x-2)(2x+1)=0
The product of two numbers or expression cannot be zero,unless one of them is zero
Therefore either x-2=0 or 2x+1=0
If x-2=0,then x=2
If 2x+1=0,then 2x= -1 or x= -1/2
x=2 or -1/2

Answer 9

first u combine like terms...2x + -3x= -1x & 2 - 2 = 0

your get is -1x =0

divide both sides by -1

0/-1=0

the answer is x=0

Answer 10

2x^2-3x-2=0 to put it simply there are two numbers for x such that when subsituted into the equation it equals zero.

2x^2-3x = 2
2x^2-3x must equal 2 for what values idk.. you solve it.. but it can be something such that expression 2x^2-3x simplifys to something like 18-16 which equals 2...someday you'll understand :D