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

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

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

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

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

x = 1, x = 2

Solution steps

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

Expand

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

x^{2} + 6 x - 3 = 3 x^{2} + 1

Move

1

to the left side

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

Move

3 x^{2}

to the left side

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

Solve with the quadratic formula

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

6^{2} - 4 (- 2) (- 4) = 2

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 1, x = 2

(x + 4)^{2} + 6 = 7 (x + 4)

- 4.9 t^{2} + 13 t + 18 = 0

5 x^{2} - 2 = - 88

25 x^{2} + 1 = 20 x

\frac{1}{2} c^{2} + 3 c = 1

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