What are the solutions to the equation (x-3)^2-(2x+5)^2=-26x-28 ?

The solutions to the equation (x-3)^2-(2x+5)^2=-26x-28 are x=2,x=-2

Find the zeros of (x-3)^2-(2x+5)^2=-26x-28

The zeros of (x-3)^2-(2x+5)^2=-26x-28 are x=2,x=-2

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(x - 3)^{2} - (2 x + 5)^{2} = - 26 x - 28

x = 2, x = - 2

Solution steps

(x - 3)^{2} - (2 x + 5)^{2} = - 26 x - 28

Expand

(x - 3)^{2} - (2 x + 5)^{2} : - 3 x^{2} - 26 x - 16

- 3 x^{2} - 26 x - 16 = - 26 x - 28

Move

16

to the right side

- 3 x^{2} - 26 x = - 26 x - 12

Move

26 x

to the left side

- 3 x^{2} = - 12

Divide both sides by

- 3

x^{2} = 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

x = 4, x = - 4

4 = 2

- 4 = - 2

x = 2, x = - 2

c^{2} - 9 c + 11 = 0

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

8 f^{2} + 11 f = 2 f^{2} + 3 (2 f - 5) + 21

y, 3 y - 2 x y - 3 y^{2} = 0

(2 x - 1) (x + 3) = - 2 x

What are the solutions to the equation (x-3)^2-(2x+5)^2=-26x-28 ?
The solutions to the equation (x-3)^2-(2x+5)^2=-26x-28 are x=2,x=-2
Find the zeros of (x-3)^2-(2x+5)^2=-26x-28
The zeros of (x-3)^2-(2x+5)^2=-26x-28 are x=2,x=-2