What are the solutions to the equation (2x-5)(x+2)-(x+2)(3x-1)=0 ?

The solutions to the equation (2x-5)(x+2)-(x+2)(3x-1)=0 are x=-4,x=-2

Find the zeros of (2x-5)(x+2)-(x+2)(3x-1)=0

The zeros of (2x-5)(x+2)-(x+2)(3x-1)=0 are x=-4,x=-2

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(2 x - 5) (x + 2) - (x + 2) (3 x - 1) = 0

x = - 4, x = - 2

Solution steps

(2 x - 5) (x + 2) - (x + 2) (3 x - 1) = 0

Expand

(2 x - 5) (x + 2) - (x + 2) (3 x - 1) : - x^{2} - 6 x - 8

- x^{2} - 6 x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 ) ( - 8 )}{2 ( - 1 )}

(- 6)^{2} - 4 (- 1) (- 8) = 2

x_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2}{2 ( - 1 )}, x_{2} = \frac{- ( - 6 ) - 2}{2 ( - 1 )}

x = \frac{- ( - 6 ) + 2}{2 ( - 1 )} : - 4

x = \frac{- ( - 6 ) - 2}{2 ( - 1 )} : - 2

The solutions to the quadratic equation are:

x = - 4, x = - 2

3 t^{2} - 40 t + 128 = 0

4 x^{2} + 36 x + 27 = 7 x^{2} + 54 x + 51

6 n^{2} = 2

x^{2} + 4 \times (- 21) = 0

4 y^{2} = - 4 y - 1

What are the solutions to the equation (2x-5)(x+2)-(x+2)(3x-1)=0 ?
The solutions to the equation (2x-5)(x+2)-(x+2)(3x-1)=0 are x=-4,x=-2
Find the zeros of (2x-5)(x+2)-(x+2)(3x-1)=0
The zeros of (2x-5)(x+2)-(x+2)(3x-1)=0 are x=-4,x=-2