What are the solutions to the equation x^2+(x+3)^2-89=0 ?

The solutions to the equation x^2+(x+3)^2-89=0 are x=5,x=-8

Find the zeros of x^2+(x+3)^2-89=0

The zeros of x^2+(x+3)^2-89=0 are x=5,x=-8

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x^{2} + (x + 3)^{2} - 89 = 0

x = 5, x = - 8

Solution steps

x^{2} + (x + 3)^{2} - 89 = 0

Expand

x^{2} + (x + 3)^{2} - 89 : 2 x^{2} + 6 x - 80

2 x^{2} + 6 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 2 ( - 80 )}{2 \cdot 2}

6^{2} - 4 \cdot 2 (- 80) = 26

x_{1, 2} = \frac{- 6 \pm 26}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 6 + 26}{2 \cdot 2}, x_{2} = \frac{- 6 - 26}{2 \cdot 2}

x = \frac{- 6 + 26}{2 \cdot 2} : 5

x = \frac{- 6 - 26}{2 \cdot 2} : - 8

The solutions to the quadratic equation are:

x = 5, x = - 8

w^{2} - w - 6 = 0

(a^{- 2} + b^{- 2})^{- 1}

2 x^{2} = 10 x

5^{4} \cdot 5^{3}

a \frac{4}{7}

What are the solutions to the equation x^2+(x+3)^2-89=0 ?
The solutions to the equation x^2+(x+3)^2-89=0 are x=5,x=-8
Find the zeros of x^2+(x+3)^2-89=0
The zeros of x^2+(x+3)^2-89=0 are x=5,x=-8