What are the solutions to the equation x^2+(2x+3)^2=26 ?

The solutions to the equation x^2+(2x+3)^2=26 are x=1,x=-17/5

Find the zeros of x^2+(2x+3)^2=26

The zeros of x^2+(2x+3)^2=26 are x=1,x=-17/5

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x^{2} + (2 x + 3)^{2} = 26

x = 1, x = - \frac{17}{5}

Decimal Notation

x = 1, x = - 3.4

Solution steps

x^{2} + (2 x + 3)^{2} = 26

Expand

x^{2} + (2 x + 3)^{2} : 5 x^{2} + 12 x + 9

5 x^{2} + 12 x + 9 = 26

Move

26

to the left side

5 x^{2} + 12 x - 17 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 5 ( - 17 )}{2 \cdot 5}

1 2^{2} - 4 \cdot 5 (- 17) = 22

x_{1, 2} = \frac{- 12 \pm 22}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 12 + 22}{2 \cdot 5}, x_{2} = \frac{- 12 - 22}{2 \cdot 5}

x = \frac{- 12 + 22}{2 \cdot 5} : 1

x = \frac{- 12 - 22}{2 \cdot 5} : - \frac{17}{5}

The solutions to the quadratic equation are:

x = 1, x = - \frac{17}{5}

x (x - 4) (x + 5) \leq 0

12 - x = 4

9 x^{2} - 24 x^{5}

\frac{11 a ^{4}}{6} \cdot \frac{12 b}{a ^{5}}

y^{2} + 8 y + 7 = 0

What are the solutions to the equation x^2+(2x+3)^2=26 ?
The solutions to the equation x^2+(2x+3)^2=26 are x=1,x=-17/5
Find the zeros of x^2+(2x+3)^2=26
The zeros of x^2+(2x+3)^2=26 are x=1,x=-17/5