What are the solutions to the equation (2x-1)(3x+2)=x+292 ?

The solutions to the equation (2x-1)(3x+2)=x+292 are x=7,x=-7

Find the zeros of (2x-1)(3x+2)=x+292

The zeros of (2x-1)(3x+2)=x+292 are x=7,x=-7

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(2 x - 1) (3 x + 2) = x + 292

x = 7, x = - 7

Solution steps

(2 x - 1) (3 x + 2) = x + 292

Expand

(2 x - 1) (3 x + 2) : 6 x^{2} + x - 2

6 x^{2} + x - 2 = x + 292

Move

2

to the right side

6 x^{2} + x = x + 294

Move

x

to the left side

6 x^{2} = 294

Divide both sides by

6

x^{2} = 49

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 49, x = - 49

49 = 7

- 49 = - 7

x = 7, x = - 7

4 a^{2} - 28 ab + 49 b^{2}

77 - 5175

4 x + 6 < 22

x^{5} y - x y^{4}

(x + 3) (x + 4) + y (x + 4)

What are the solutions to the equation (2x-1)(3x+2)=x+292 ?
The solutions to the equation (2x-1)(3x+2)=x+292 are x=7,x=-7
Find the zeros of (2x-1)(3x+2)=x+292
The zeros of (2x-1)(3x+2)=x+292 are x=7,x=-7