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

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

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

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

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

x = \frac{6 - 6}{3}, x = \frac{6 + 6}{3}

Decimal Notation

x = 1.18350 \dots, x = 2.81649 \dots

Solution steps

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

Expand

x^{2} - (2 x - 3)^{2} : - 3 x^{2} + 12 x - 9

- 3 x^{2} + 12 x - 9 = 1

Move

1

to the left side

- 3 x^{2} + 12 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) ( - 10 )}{2 ( - 3 )}

1 2^{2} - 4 (- 3) (- 10) = 26

x_{1, 2} = \frac{- 12 \pm 2 6}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 12 + 2 6}{2 ( - 3 )}, x_{2} = \frac{- 12 - 2 6}{2 ( - 3 )}

x = \frac{- 12 + 2 6}{2 ( - 3 )} : \frac{6 - 6}{3}

x = \frac{- 12 - 2 6}{2 ( - 3 )} : \frac{6 + 6}{3}

The solutions to the quadratic equation are:

x = \frac{6 - 6}{3}, x = \frac{6 + 6}{3}

\frac{2}{3} x^{2} + 30 = 0

- 5 = - \frac{7}{3} (7) + b

(x - 3)^{2} - 3 x + 9

7 A + 5 \geq 40

\frac{1}{7} - 3 (\frac{3}{7} n - \frac{2}{7})

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