# Please solve the equation: (2x+1)(3x-2)=55

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### 4 Answers

We have to solve (2x+1)(3x-2) = 55

Now (2x+1)(3x-2) = 55

=> 6x^2 + 3x - 4x - 2 = 55

=> 6x^2 - x - 2 = 55

=> 6x^2 - x - 57 = 0

Now the roots of the equation 6x^2 - x - 57 = 0 are

x1 = [1 + sqrt(1 + 6*4*57)] / 12

=> x1 = (1 + sqrt 1369)/12

=> x1 = (1 + 37) /12

=>x1 = 38 /12

=> x1 = 19/6

x2 = [1 - sqrt(1 + 6*4*57)] / 12

=> x2 = -36 /12

=> x2 = -3

**Therefore the roots are -3 and 19/6.**

(2x+ 1)(3x-2) = 55

We need to find x values that satisfies the equation.

First, let us expand the brackets.

==> 2x*3x + 1*3x + 2x*-2 + 1*-2 = 55

==> 6x^2 +3x - 4x - 2 = 55

Now we will subtract 55 from both sides.

==> 6x^2 - x -2 - 55 = 0

==> 6x^2 - x - 57 = 0.

Now we can factor or use the formula to determine the roots of the quadratic equation.

Let us factor.

==> (6x-19) ( x +3) = 0

==> (6x-19) = 0 ==> 6x = 19 ==> x = 19/6

==> (x+3) = 0 ==> x = -3.

Then, there are two possible solutions.

**x = { -3, 19/6}**

To solve the equation (2x+1)(3x-2) = 55.

We expand the left side and rewrite the equation:

2x(3x-2)+1(3x-2) = 55.

6x^2-4x+3x-2 = 55.

6x^2--x-2-55 = 0.

6x^2-x-57 = 0.

6x^2- 19x+18x -57 = 0.

(6x-19) +3(6x-19) = 0.

(6x-19) (x+3) = 0.

6x-19 = 0, or x+3 = 0.

6x= 19 , so x= 19/6.

x+3 = 0, so x= -3.

Therefore x= -3 and x = 19/6.

First, we'll remove the brackets:

(2x+1)(3x-2)=55

2x(3x-2) + 1*(3x-2) = 55

2x*3x - 2x*2 + 1*3x - 1*2 = 55

6x^2 - 4x + 3x - 2 = 55

We'll combine middle terms:

6x^2 - x - 2 = 55

We'll subtract 55 both sides:

6x^2 - x - 2 - 55 = 0

6x^2 - x - 57 = 0

We'll apply the quadratic formula:

x1 = [1 + sqrt(1 + 24*57)]/12

x1 = (1 + sqrt1369)/12

x1 = (1 + 37)/12

x1 = 38/12

x1 = 19/6

x2 = (1-37)/12

x2 = -36/12

x2 = -3

**The solutions of the equation are: {-3 ; 19/6}.**