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

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

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

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

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(2 - x) (2 + x) = 1

x = - 3, x = 3

Solution steps

(2 - x) (2 + x) = 1

Expand

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

4 - x^{2} = 1

Move

1

to the left side

- x^{2} + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

0^{2} - 4 (- 1) \cdot 3 = 23

x_{1, 2} = \frac{- 0 \pm 2 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 0 + 2 3}{2 ( - 1 )}, x_{2} = \frac{- 0 - 2 3}{2 ( - 1 )}

x = \frac{- 0 + 2 3}{2 ( - 1 )} : - 3

x = \frac{- 0 - 2 3}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = - 3, x = 3

2 x^{2} = 19 x + 33

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

14 x^{2} - 23 x + 15 = 12

4 t^{2} + 17 t + 18 = t^{2}

5 a^{2} - 15 a + 11 = 0

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