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

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

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

The zeros of (x-2)^2=-3x+6 are x=2,x=-1

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(x - 2)^{2} = - 3 x + 6

x = 2, x = - 1

Solution steps

(x - 2)^{2} = - 3 x + 6

Expand

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

x^{2} - 4 x + 4 = - 3 x + 6

Move

6

to the left side

x^{2} - 4 x - 2 = - 3 x

Move

3 x

to the left side

x^{2} - x - 2 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 1 \cdot (- 2) = 3

x_{1, 2} = \frac{- ( - 1 ) \pm 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 3}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 3}{2 \cdot 1}

x = \frac{- ( - 1 ) + 3}{2 \cdot 1} : 2

x = \frac{- ( - 1 ) - 3}{2 \cdot 1} : - 1

The solutions to the quadratic equation are:

x = 2, x = - 1

x^{2} + 8 x + 7

5 x^{2} + 2 x = 7

5^{2} = x^{2} + 3^{2}

12 x^{2} - 8 x = 0

m^{2} + 4 m - 5 = 0

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