# Solve for x: (3x+7)(x-1)=24.

hala718 | Certified Educator

(3x+7)(x-1) = 24

We will expand the brackets.

==> 3x*x + 7*x + 3x*-1 + 7*-1 = 24

==> 3x^2 + 7x-3x - 7 = 24

We will combine like terms.

==> 3x^2+ 4x -7 - 24 = 0

==> 3x^2 + 4x - 31 = 0

Now we will use the formula to find the roots of the quadratic equation.

==> x1= ( -4 + sqrt(16+4*3*31) / 6

= ( -4 + sqrt(388) / 6 = [ -4 + 2sqrt(97)] / 6

==> x1= -2/3 + sqrt97 / 3

==> x2= -2/3 - sqrt97 / 3

giorgiana1976 | Student

First, we'll remove the brackets:

(3x+7)(x-1)=24

3x(x-1) + 7*(x-1) = 24

3x*x - 3x*1 + 7*x - 7*1 = 24

3x^2 - 3x + 7x - 7 = 24

We'll combine middle terms:

3x^2 + 4x - 7 = 24

We'll subtract 24 both sides:

3x^2 + 4x - 31 = 0

x1 = [-4 + sqrt(16 + 372)]/6

x1 = (-4+ 2sqrt97)/6

x1 = (-2 + sqrt97)/3

x2 = (-2 - sqrt97)/3

The solutions of the equation are: {(-2 - sqrt97)/3 ; (-2 + sqrt97)/3}.

neela | Student

To solve for x: (3x+7)(x-1)=24.

We expand left  and rewrite the equation:

3x(x-1)+7(x-1) = 24.

3x^2-3x+7x-7 = 24.

=> 3x^2 +4x-7-24 = 0.

=> 3x^2 +4x-31 = 0....(1).

This is in the form of ax^2+bx+c = 0 whose solution is given by:

x1 = (-b+sqrt(b^2-4ac)}/2a and

x2 = (-b-sqrt(b^2-4ac)}/2a.

Here in equation at (1): a = 3, b= 4 and c = -31.

Therefore x1 = (-4+sqrt(4^2-4*3*-31)}/2*3 = (-4+sqrt388)/6

=> x1 = (-2+sqrt97)/3

x2 = (-2-sqrt97)/3.

Therefore  (-2+sqrt97)/3 and (-2-sqrt97)/3 are the solutions.