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

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.

Approved by eNotes Editorial Team (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}

Approved by eNotes Editorial Team