What are the solutions to the equation (x-3)^2-(2x+5)^2=-16 ?

The solutions to the equation (x-3)^2-(2x+5)^2=-16 are x=-26/3 ,x=0

Find the zeros of (x-3)^2-(2x+5)^2=-16

The zeros of (x-3)^2-(2x+5)^2=-16 are x=-26/3 ,x=0

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

x = - \frac{26}{3}, x = 0

Decimal Notation

x = - 8.66666 \dots, x = 0

Solution steps

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

Expand

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

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

Move

16

to the left side

- 3 x^{2} - 26 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 26 ) \pm ( - 26 ) ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

(- 26)^{2} - 4 (- 3) \cdot 0 = 26

x_{1, 2} = \frac{- ( - 26 ) \pm 26}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- ( - 26 ) + 26}{2 ( - 3 )}, x_{2} = \frac{- ( - 26 ) - 26}{2 ( - 3 )}

x = \frac{- ( - 26 ) + 26}{2 ( - 3 )} : - \frac{26}{3}

x = \frac{- ( - 26 ) - 26}{2 ( - 3 )} : 0

The solutions to the quadratic equation are:

x = - \frac{26}{3}, x = 0

12 - x^{2} = - 3

x, x^{2} + x y + y^{2} = 3

4 x^{2} + 8 x + 13 = 0

12 x - 7 x^{2} + 64 = 0

4 x^{2} + 27 = x^{2}

What are the solutions to the equation (x-3)^2-(2x+5)^2=-16 ?
The solutions to the equation (x-3)^2-(2x+5)^2=-16 are x=-26/3 ,x=0
Find the zeros of (x-3)^2-(2x+5)^2=-16
The zeros of (x-3)^2-(2x+5)^2=-16 are x=-26/3 ,x=0