What are the solutions to the equation (x+2)(x-3)-x(2x+1)+6x=0 ?

The solutions to the equation (x+2)(x-3)-x(2x+1)+6x=0 are x=2-sqrt(2)i,x=2+sqrt(2)i

Find the zeros of (x+2)(x-3)-x(2x+1)+6x=0

The zeros of (x+2)(x-3)-x(2x+1)+6x=0 are x=2-sqrt(2)i,x=2+sqrt(2)i

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(x + 2) (x - 3) - x (2 x + 1) + 6 x = 0

x = 2 - 2 i, x = 2 + 2 i

Solution steps

(x + 2) (x - 3) - x (2 x + 1) + 6 x = 0

Expand

(x + 2) (x - 3) - x (2 x + 1) + 6 x : - x^{2} + 4 x - 6

- x^{2} + 4 x - 6 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 (- 1) (- 6) : 22 i

x_{1, 2} = \frac{- 4 \pm 2 2 i}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 2 2 i}{2 ( - 1 )}, x_{2} = \frac{- 4 - 2 2 i}{2 ( - 1 )}

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

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

The solutions to the quadratic equation are:

x = 2 - 2 i, x = 2 + 2 i

4 x^{2} - 2 = 94

5 x^{2} - 125 x = 0

p^{2} + 2 p - 8 = 0

x^{2} + 5 x + \frac{21}{4} = 0

x^{2} - 3 = 4

What are the solutions to the equation (x+2)(x-3)-x(2x+1)+6x=0 ?
The solutions to the equation (x+2)(x-3)-x(2x+1)+6x=0 are x=2-sqrt(2)i,x=2+sqrt(2)i
Find the zeros of (x+2)(x-3)-x(2x+1)+6x=0
The zeros of (x+2)(x-3)-x(2x+1)+6x=0 are x=2-sqrt(2)i,x=2+sqrt(2)i