What are the solutions to the equation (x-2)^2+(x+3)^2=25 ?

The solutions to the equation (x-2)^2+(x+3)^2=25 are x=2,x=-3

Find the zeros of (x-2)^2+(x+3)^2=25

The zeros of (x-2)^2+(x+3)^2=25 are x=2,x=-3

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(x - 2)^{2} + (x + 3)^{2} = 25

x = 2, x = - 3

Solution steps

(x - 2)^{2} + (x + 3)^{2} = 25

Expand

(x - 2)^{2} + (x + 3)^{2} : 2 x^{2} + 2 x + 13

2 x^{2} + 2 x + 13 = 25

Move

25

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 2 (- 12) = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 \cdot 2}, x_{2} = \frac{- 2 - 10}{2 \cdot 2}

x = \frac{- 2 + 10}{2 \cdot 2} : 2

x = \frac{- 2 - 10}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

x = 2, x = - 3

x^{2} - 5 x = 0

(- \frac{3}{4})^{2} + y^{2} = 1

- 3 x^{2} - 7 x + 2 = 6

0 = 4 x^{2} - 8 x - 5

D, D^{2} y - Dy - 6 y = e^{2 x}

What are the solutions to the equation (x-2)^2+(x+3)^2=25 ?
The solutions to the equation (x-2)^2+(x+3)^2=25 are x=2,x=-3
Find the zeros of (x-2)^2+(x+3)^2=25
The zeros of (x-2)^2+(x+3)^2=25 are x=2,x=-3