What are the solutions to the equation x^2+(x+21)^2=51^2 ?

The solutions to the equation x^2+(x+21)^2=51^2 are x=24,x=-45

Find the zeros of x^2+(x+21)^2=51^2

The zeros of x^2+(x+21)^2=51^2 are x=24,x=-45

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x^{2} + (x + 21)^{2} = 5 1^{2}

x = 24, x = - 45

Solution steps

x^{2} + (x + 21)^{2} = 5 1^{2}

Expand

x^{2} + (x + 21)^{2} : 2 x^{2} + 42 x + 441

Expand

5 1^{2} : 2601

2 x^{2} + 42 x + 441 = 2601

Move

2601

to the left side

2 x^{2} + 42 x - 2160 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 42 \pm 4 2 ^{2} - 4 \cdot 2 ( - 2160 )}{2 \cdot 2}

4 2^{2} - 4 \cdot 2 (- 2160) = 138

x_{1, 2} = \frac{- 42 \pm 138}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 42 + 138}{2 \cdot 2}, x_{2} = \frac{- 42 - 138}{2 \cdot 2}

x = \frac{- 42 + 138}{2 \cdot 2} : 24

x = \frac{- 42 - 138}{2 \cdot 2} : - 45

The solutions to the quadratic equation are:

x = 24, x = - 45

7 x^{2} = 25

n^{2} - 9 n + 8

- 4 x^{2} + 36 = 0

w^{2} - 7 w + 19 = 0

- 2 w^{2} - 12 w - 9 = 0

What are the solutions to the equation x^2+(x+21)^2=51^2 ?
The solutions to the equation x^2+(x+21)^2=51^2 are x=24,x=-45
Find the zeros of x^2+(x+21)^2=51^2
The zeros of x^2+(x+21)^2=51^2 are x=24,x=-45