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

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

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

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

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x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} = 0

x = 1, x = 3

Solution steps

x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} = 0

Expand

x^{2} + 2 x (- x + 2) - 3 (- x + 2)^{2} : - 4 x^{2} + 16 x - 12

- 4 x^{2} + 16 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 4 ) ( - 12 )}{2 ( - 4 )}

1 6^{2} - 4 (- 4) (- 12) = 8

x_{1, 2} = \frac{- 16 \pm 8}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 16 + 8}{2 ( - 4 )}, x_{2} = \frac{- 16 - 8}{2 ( - 4 )}

x = \frac{- 16 + 8}{2 ( - 4 )} : 1

x = \frac{- 16 - 8}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

x = 1, x = 3

5 x^{2} = 6 x - 1

3 u^{2} + 24 u = 0

16 x^{2} = - 56 x - 18

x^{2} + 5 x + 1 = 0

t, - 4.905 t^{2} - 0.517 t - 15 = 0

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