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

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

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

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

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3 x^{2} = 2 (x - 1)^{2} + 3

x = - 5, x = 1

Solution steps

3 x^{2} = 2 (x - 1)^{2} + 3

Expand

2 (x - 1)^{2} + 3 : 2 x^{2} - 4 x + 5

3 x^{2} = 2 x^{2} - 4 x + 5

Switch sides

2 x^{2} - 4 x + 5 = 3 x^{2}

Move

3 x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 1 ) \cdot 5}{2 ( - 1 )}

(- 4)^{2} - 4 (- 1) \cdot 5 = 6

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

Separate the solutions

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

x = \frac{- ( - 4 ) + 6}{2 ( - 1 )} : - 5

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

The solutions to the quadratic equation are:

x = - 5, x = 1

x^{(2)} + 20 x + 28 = 9

6 - 5 x + x^{2} = 0

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

2 t^{2} + 4 t + 2 = 0

x^{2} - 2 x - 22 = 2 x - 1

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