What are the solutions to the equation x^2+(2x+1)^2=10 ?

The solutions to the equation x^2+(2x+1)^2=10 are x=1,x=-9/5

Find the zeros of x^2+(2x+1)^2=10

The zeros of x^2+(2x+1)^2=10 are x=1,x=-9/5

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x^{2} + (2 x + 1)^{2} = 10

x = 1, x = - \frac{9}{5}

Decimal Notation

x = 1, x = - 1.8

Solution steps

x^{2} + (2 x + 1)^{2} = 10

Expand

x^{2} + (2 x + 1)^{2} : 5 x^{2} + 4 x + 1

5 x^{2} + 4 x + 1 = 10

Move

10

to the left side

5 x^{2} + 4 x - 9 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 5 ( - 9 )}{2 \cdot 5}

4^{2} - 4 \cdot 5 (- 9) = 14

x_{1, 2} = \frac{- 4 \pm 14}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 4 + 14}{2 \cdot 5}, x_{2} = \frac{- 4 - 14}{2 \cdot 5}

x = \frac{- 4 + 14}{2 \cdot 5} : 1

x = \frac{- 4 - 14}{2 \cdot 5} : - \frac{9}{5}

The solutions to the quadratic equation are:

x = 1, x = - \frac{9}{5}

5 n^{2} = 6 n - 1

- 3 = g^{2} - 4 g

- x^{2} + 2 x = - 1

2 y^{2} = 16

2 y^{2} = 54

What are the solutions to the equation x^2+(2x+1)^2=10 ?
The solutions to the equation x^2+(2x+1)^2=10 are x=1,x=-9/5
Find the zeros of x^2+(2x+1)^2=10
The zeros of x^2+(2x+1)^2=10 are x=1,x=-9/5