What are the solutions to the equation x^2+2(4)x-3=(4)x^2+1 ?

The solutions to the equation x^2+2(4)x-3=(4)x^2+1 are x= 2/3 ,x=2

Find the zeros of x^2+2(4)x-3=(4)x^2+1

The zeros of x^2+2(4)x-3=(4)x^2+1 are x= 2/3 ,x=2

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

x = \frac{2}{3}, x = 2

Decimal Notation

x = 0.66666 \dots, x = 2

Solution steps

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

Expand

x^{2} + 2 (4) x - 3 : x^{2} + 8 x - 3

x^{2} + 8 x - 3 = 4 x^{2} + 1

Move

1

to the left side

x^{2} + 8 x - 4 = 4 x^{2}

Move

4 x^{2}

to the left side

- 3 x^{2} + 8 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 3 ) ( - 4 )}{2 ( - 3 )}

8^{2} - 4 (- 3) (- 4) = 4

x_{1, 2} = \frac{- 8 \pm 4}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 8 + 4}{2 ( - 3 )}, x_{2} = \frac{- 8 - 4}{2 ( - 3 )}

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

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

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = 2

6 x - 0.06 x^{2} = 0

0 = 3 x^{2} + 10 x + 180

y^{2} = 2 y + 15

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

x^{2} + 4 x - 60 = 0

What are the solutions to the equation x^2+2(4)x-3=(4)x^2+1 ?
The solutions to the equation x^2+2(4)x-3=(4)x^2+1 are x= 2/3 ,x=2
Find the zeros of x^2+2(4)x-3=(4)x^2+1
The zeros of x^2+2(4)x-3=(4)x^2+1 are x= 2/3 ,x=2