What are the solutions to the equation (x-2)^2+(x+1)^2=49 ?

The solutions to the equation (x-2)^2+(x+1)^2=49 are x=(1+sqrt(89))/2 ,x=(1-sqrt(89))/2

Find the zeros of (x-2)^2+(x+1)^2=49

The zeros of (x-2)^2+(x+1)^2=49 are x=(1+sqrt(89))/2 ,x=(1-sqrt(89))/2

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(x - 2)^{2} + (x + 1)^{2} = 49

x = \frac{1 + 89}{2}, x = \frac{1 - 89}{2}

Solution steps

(x - 2)^{2} + (x + 1)^{2} = 49

Expand

(x - 2)^{2} + (x + 1)^{2} : 2 x^{2} - 2 x + 5

2 x^{2} - 2 x + 5 = 49

Move

49

to the left side

2 x^{2} - 2 x - 44 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 2 ( - 44 )}{2 \cdot 2}

(- 2)^{2} - 4 \cdot 2 (- 44) = 289

x_{1, 2} = \frac{- ( - 2 ) \pm 2 89}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 89}{2 \cdot 2}, x_{2} = \frac{- ( - 2 ) - 2 89}{2 \cdot 2}

x = \frac{- ( - 2 ) + 2 89}{2 \cdot 2} : \frac{1 + 89}{2}

x = \frac{- ( - 2 ) - 2 89}{2 \cdot 2} : \frac{1 - 89}{2}

The solutions to the quadratic equation are:

x = \frac{1 + 89}{2}, x = \frac{1 - 89}{2}

\frac{6 - 3 5}{2 + 5}

\frac{s ^{2} t ^{- 4}}{5 s ^{- 1} t}

∣ 5 x - 2 ∣ + 4 = 72

36 c^{2} - d^{2}

12 a = - 192

What are the solutions to the equation (x-2)^2+(x+1)^2=49 ?
The solutions to the equation (x-2)^2+(x+1)^2=49 are x=(1+sqrt(89))/2 ,x=(1-sqrt(89))/2
Find the zeros of (x-2)^2+(x+1)^2=49
The zeros of (x-2)^2+(x+1)^2=49 are x=(1+sqrt(89))/2 ,x=(1-sqrt(89))/2