how do you Solve this equation 2x^2+3x=(2x-1)(x+7)?

Answer 1

Here you go:

1) 2x^2+3x=(2x-1)(x+7)
multiply out (2x-1)(x+7) to get:

2) 2x^2+3x=2x^2+14x-x-7
subtract the 2x^2 from each side and subtract 14x-x to get:

3) 3x = 13x-7
Solve for x to get:

4) x = 7/10

You can put 7/10 back in the equation on each side and get 3.08 = 3.08.

Good Luck!

Answer 2

2x^2 + 3x = (2x - 1)(x + 7)

2x^2 + 3x = 2x^2 + 14 -x - 7
(cancel 2x^2 on both sides of the equation)

11x - x - 7 = 0

10x = 7

x = 7/10

Answer 3

ok i'm answering from a iPhone so bare with me. Use the foil technique and paintings the equation backwards actual. ( 2x - 3 ) (x +4 ) = 0 in case you prolonged this out you get your unique equation so it is applicable. ok so now set each factor equivalent to 0. ( 2x - 3) = 0. upload 3 to the two factors and your equation reads 2x = 3. purely sparkling up for x now. Divide the two factors by way of 2 and x will equivalent 3/2 that's one answer. Now set the different factor equivalent to 0. Your equation will examine x + 4 = 0 . purely subtract 4 from the two factors and you get x = -4 your 2 strategies are x= 3/2 and x= -4

Answer 4

2x^2 + 3x = (2x - 1) (x+7)
2x^2 +3x = 2x^2 +14x - x - 7
2x^2 + 3x = 2x^2 +13x - 7
simplify by transposition;
2x^2 - 2x^2 +3x -13x = -7
-10x = -7
-10x/10= -7/10
-x = -7/10
x= 7/10

Answer 5

2x^2+3x=2x^2+14x-1x-7
we get this equation by opening bracket
2x^2+3x=2x^2+13x-7
2x^2-2x^2+3x-13x= -7
0-10x=-7
Value of x= 7/10

Answer 6

2x^2 + 3x = (2x-1) (x+7)
=> 2x^2 + 3x = 2x^2 + 14x -x -7
=> 10x = 7
=> x = 0.7

Answer 7

2x² + 3x= (2x-1)(x+7)

2x² + 3x = 2x² + 13x - 7

Combine terms

3x = 13x - 7

-10x = -7

x = 7/10 or 0.7
.

Answer 8

You can't solve it.....
You just simplified the equation.....
that's all