(3x-5)(x+2) = -6?

Question

pre algebra problem. i supposed use one of the quadratic method to solve the equation. please help thank you

Answer 1

3x² + x - 10 = - 6
3x² + x - 4 = 0
x = [- 1 ± √(1 + 48) ] / 6
x = [- 1 ± √(49) ] / 6
x = [- 1 ± 7 ] / 6
x = 1 , x = - 4 / 3

Answer 2

You have to multiply the factors, incorporate the constant term (first responder did not do this), then either re-factor or use the quadratic equation. We get 3x^2 +x -10 + 6 = 0, or 3x^2 +x -4 = 0, which factors to get (3x +4)(x -1) = 0. So the roots are -4/3 and 1.

Answer 3

Above isn't correct.

3x^2 + 6x -5x - 10 = -6
3x^2 + 6x -5x - 10 + 6 = 0
3x^2 + x - 4 = 0

Next:
3x^2 + x - 4 = 0
3x^2 + 4x - 3x - 4 = 0
(3x+4)(x-1)= 0
x = -4/3 , 1

OR

Use the quadratic formula, a = 3, b =1, etc.

Answer 4

(3x-5)(x+2) = -6............Given Equation

Distribute the first element & second element of the first bracket to all the elements of second bracket .....we have

3x (x + 2) -5(x + 2) = -6

3x^2 + 6x -5x -10 = -6

Group the common ( x- terms)....

3x^2 + x -10 = -6 ( since +6x - 5x = x or

Group the number terms like ( -10 & -6).....we have

3x^2 + x = - 6 +10 ( this step is got by adding 10 on both sides)

3x^2 + x = 4

Now subtracting 4 on both sides, we have

3x^2 + x - 4 = 4 - 4

3x^2 + x - 4 = 0 ...............Final Answer


Regards,
Abhishek ( Algebra Tutor- Eduwizards )
www.eduwizards.com

Answer 5

(3x-5)(x+2) = -6 (Expand it)
3x2 + 6x – 5x –10 = -6
3x2 + x – 10 + 6 = 0 (Move -6 over)
3x2 + x – 4 = 0
(3x + 4) (x – 1) = 0
3x + 4 = 0 or x – 1 = 0
x = -4/3 x = 1

Answer 6

(3x-5)(x+2)=-6
opening the bracket results to:
3x^2+x-10=-6
3x^2+x-4=0
if u factor it:
3x^2-3x+4x-4=0
3x(x-1)+4(x-1)=0
(3x+4)(x-1)=0
3x+4=0;x-1=0
3x=-4;x=1
x=-4/3,x=1

Answer 7

(3x-5) = -6 ; (x+2) = -6
3x = -6 +5 ; x = -6 -2
3x = -1 ; x = -8
x = -1/3 ; x = -8