What are the solutions to the equation (2x+3)^2+8(2x+3)+11=0 ?

The solutions to the equation (2x+3)^2+8(2x+3)+11=0 are x=(-7+sqrt(5))/2 ,x=-(7+sqrt(5))/2

Find the zeros of (2x+3)^2+8(2x+3)+11=0

The zeros of (2x+3)^2+8(2x+3)+11=0 are x=(-7+sqrt(5))/2 ,x=-(7+sqrt(5))/2

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(2 x + 3)^{2} + 8 (2 x + 3) + 11 = 0

x = \frac{- 7 + 5}{2}, x = - \frac{7 + 5}{2}

Decimal Notation

x = - 2.38196 \dots, x = - 4.61803 \dots

Solution steps

(2 x + 3)^{2} + 8 (2 x + 3) + 11 = 0

Expand

(2 x + 3)^{2} + 8 (2 x + 3) + 11 : 4 x^{2} + 28 x + 44

4 x^{2} + 28 x + 44 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 4 \cdot 44}{2 \cdot 4}

2 8^{2} - 4 \cdot 4 \cdot 44 = 45

x_{1, 2} = \frac{- 28 \pm 4 5}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 28 + 4 5}{2 \cdot 4}, x_{2} = \frac{- 28 - 4 5}{2 \cdot 4}

x = \frac{- 28 + 4 5}{2 \cdot 4} : \frac{- 7 + 5}{2}

x = \frac{- 28 - 4 5}{2 \cdot 4} : - \frac{7 + 5}{2}

The solutions to the quadratic equation are:

x = \frac{- 7 + 5}{2}, x = - \frac{7 + 5}{2}

15 x^{2} + 14 x - 16

- 4 (x + 5) + 3 x - 1 = - 13 - 13 - 13

- 4 \leq - \frac{2}{3} v + 2

3 k^{2} + 13 k - 56

48 m - 80 n

What are the solutions to the equation (2x+3)^2+8(2x+3)+11=0 ?
The solutions to the equation (2x+3)^2+8(2x+3)+11=0 are x=(-7+sqrt(5))/2 ,x=-(7+sqrt(5))/2
Find the zeros of (2x+3)^2+8(2x+3)+11=0
The zeros of (2x+3)^2+8(2x+3)+11=0 are x=(-7+sqrt(5))/2 ,x=-(7+sqrt(5))/2