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

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

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

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

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

x = - 3 + 22, x = - 3 - 22

Decimal Notation

x = - 0.17157 \dots, x = - 5.82842 \dots

Solution steps

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

Expand

(2 x + 3)^{2} - (3 - x)^{2} : 3 x^{2} + 18 x

3 x^{2} + 18 x = - 3

Move

3

to the left side

3 x^{2} + 18 x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

1 8^{2} - 4 \cdot 3 \cdot 3 = 122

x_{1, 2} = \frac{- 18 \pm 12 2}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 18 + 12 2}{2 \cdot 3}, x_{2} = \frac{- 18 - 12 2}{2 \cdot 3}

x = \frac{- 18 + 12 2}{2 \cdot 3} : - 3 + 22

x = \frac{- 18 - 12 2}{2 \cdot 3} : - 3 - 22

The solutions to the quadratic equation are:

x = - 3 + 22, x = - 3 - 22

c^{2} = 4 + 25

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

x, z = 36 - 3 x^{2} - 3 y^{2}

(x - 8) (3 x + 5) = 0

b = x^{2}

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