What are the solutions to the equation x^2+(2x+3)^2=24 ?

The solutions to the equation x^2+(2x+3)^2=24 are x=(-6+sqrt(111))/5 ,x=-(6+sqrt(111))/5

Find the zeros of x^2+(2x+3)^2=24

The zeros of x^2+(2x+3)^2=24 are x=(-6+sqrt(111))/5 ,x=-(6+sqrt(111))/5

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x^{2} + (2 x + 3)^{2} = 24

x = \frac{- 6 + 111}{5}, x = - \frac{6 + 111}{5}

Decimal Notation

x = 0.90713 \dots, x = - 3.30713 \dots

Solution steps

x^{2} + (2 x + 3)^{2} = 24

Expand

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

5 x^{2} + 12 x + 9 = 24

Move

24

to the left side

5 x^{2} + 12 x - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 5 ( - 15 )}{2 \cdot 5}

1 2^{2} - 4 \cdot 5 (- 15) = 2111

x_{1, 2} = \frac{- 12 \pm 2 111}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- 12 + 2 111}{2 \cdot 5}, x_{2} = \frac{- 12 - 2 111}{2 \cdot 5}

x = \frac{- 12 + 2 111}{2 \cdot 5} : \frac{- 6 + 111}{5}

x = \frac{- 12 - 2 111}{2 \cdot 5} : - \frac{6 + 111}{5}

The solutions to the quadratic equation are:

x = \frac{- 6 + 111}{5}, x = - \frac{6 + 111}{5}

2 0^{2} + 7^{2} = x^{2}

t, e = e_{0} + v_{0} \cdot t + \frac{1}{2} \cdot a t^{2}

y, x^{2} + (y - 1)^{2} = 4

x, 2 x^{2} - 5 x y - 3 y^{2} = 0

n (x) = (x - 2)^{2} - 3

What are the solutions to the equation x^2+(2x+3)^2=24 ?
The solutions to the equation x^2+(2x+3)^2=24 are x=(-6+sqrt(111))/5 ,x=-(6+sqrt(111))/5
Find the zeros of x^2+(2x+3)^2=24
The zeros of x^2+(2x+3)^2=24 are x=(-6+sqrt(111))/5 ,x=-(6+sqrt(111))/5