What are the solutions to the equation (2+t)^2+(9+t)^2=169 ?

The solutions to the equation (2+t)^2+(9+t)^2=169 are t=3,t=-14

Find the zeros of (2+t)^2+(9+t)^2=169

The zeros of (2+t)^2+(9+t)^2=169 are t=3,t=-14

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(2 + t)^{2} + (9 + t)^{2} = 169

t = 3, t = - 14

Solution steps

(2 + t)^{2} + (9 + t)^{2} = 169

Expand

(2 + t)^{2} + (9 + t)^{2} : 2 t^{2} + 22 t + 85

2 t^{2} + 22 t + 85 = 169

Move

169

to the left side

2 t^{2} + 22 t - 84 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 \cdot 2 ( - 84 )}{2 \cdot 2}

2 2^{2} - 4 \cdot 2 (- 84) = 34

t_{1, 2} = \frac{- 22 \pm 34}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- 22 + 34}{2 \cdot 2}, t_{2} = \frac{- 22 - 34}{2 \cdot 2}

t = \frac{- 22 + 34}{2 \cdot 2} : 3

t = \frac{- 22 - 34}{2 \cdot 2} : - 14

The solutions to the quadratic equation are:

t = 3, t = - 14

x^{2} - (n - 3) x + 16 = 0

7 x = 2 x^{2} - 4

x, x^{2} p^{2} + y^{2} q^{2} = z^{2}

6 x^{2} + 5 x = 1

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

What are the solutions to the equation (2+t)^2+(9+t)^2=169 ?
The solutions to the equation (2+t)^2+(9+t)^2=169 are t=3,t=-14
Find the zeros of (2+t)^2+(9+t)^2=169
The zeros of (2+t)^2+(9+t)^2=169 are t=3,t=-14