What are the solutions to the equation 18+x(x-4)=2(x-3)^2 ?

The solutions to the equation 18+x(x-4)=2(x-3)^2 are x=0,x=8

Find the zeros of 18+x(x-4)=2(x-3)^2

The zeros of 18+x(x-4)=2(x-3)^2 are x=0,x=8

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18 + x (x - 4) = 2 (x - 3)^{2}

x = 0, x = 8

Solution steps

18 + x (x - 4) = 2 (x - 3)^{2}

Expand

18 + x (x - 4) : 18 + x^{2} - 4 x

Expand

2 (x - 3)^{2} : 2 x^{2} - 12 x + 18

18 + x^{2} - 4 x = 2 x^{2} - 12 x + 18

Move

18

to the left side

x^{2} - 4 x = 2 x^{2} - 12 x

Move

12 x

to the left side

x^{2} + 8 x = 2 x^{2}

Move

2 x^{2}

to the left side

- x^{2} + 8 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) \cdot 0}{2 ( - 1 )}

8^{2} - 4 (- 1) \cdot 0 = 8

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

Separate the solutions

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

x = \frac{- 8 + 8}{2 ( - 1 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 8

145 = 5 m^{2} - 15

x (x - 6) = - 9

3 x^{2} + 10 x - 9 = 0

x^{2} = 92

x^{2} = 73

What are the solutions to the equation 18+x(x-4)=2(x-3)^2 ?
The solutions to the equation 18+x(x-4)=2(x-3)^2 are x=0,x=8
Find the zeros of 18+x(x-4)=2(x-3)^2
The zeros of 18+x(x-4)=2(x-3)^2 are x=0,x=8