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...
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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}