Please solve the equation: (2x+1)(3x-2)=55
4 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
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.
hala718 | High School Teacher | (Level 1) Educator Emeritus
(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}
neela | High School Teacher | (Level 3) Valedictorian
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.
giorgiana1976 | College Teacher | (Level 3) Valedictorian
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}.