What are the solutions to the equation (x)^2+(x+2)^2=724 ?

The solutions to the equation (x)^2+(x+2)^2=724 are x=18,x=-20

Find the zeros of (x)^2+(x+2)^2=724

The zeros of (x)^2+(x+2)^2=724 are x=18,x=-20

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(x)^{2} + (x + 2)^{2} = 724

x = 18, x = - 20

Solution steps

x^{2} + (x + 2)^{2} = 724

Expand

x^{2} + (x + 2)^{2} : 2 x^{2} + 4 x + 4

2 x^{2} + 4 x + 4 = 724

Move

724

to the left side

2 x^{2} + 4 x - 720 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 720 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 720) = 76

x_{1, 2} = \frac{- 4 \pm 76}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 76}{2 \cdot 2}, x_{2} = \frac{- 4 - 76}{2 \cdot 2}

x = \frac{- 4 + 76}{2 \cdot 2} : 18

x = \frac{- 4 - 76}{2 \cdot 2} : - 20

The solutions to the quadratic equation are:

x = 18, x = - 20

r^{2} + 10 r - 9 = 0

9 x^{2} - 12 = 0

(2 x - 7)^{2} + 16 = 0

10 x^{2} + 4 x + 3920 = 0

2 m^{2} + 4 = 0

What are the solutions to the equation (x)^2+(x+2)^2=724 ?
The solutions to the equation (x)^2+(x+2)^2=724 are x=18,x=-20
Find the zeros of (x)^2+(x+2)^2=724
The zeros of (x)^2+(x+2)^2=724 are x=18,x=-20