What are the solutions to the equation (2x+1)^2+5(2x+1)-8=0 ?

The solutions to the equation (2x+1)^2+5(2x+1)-8=0 are x=(-7+sqrt(57))/4 ,x=-(7+sqrt(57))/4

Find the zeros of (2x+1)^2+5(2x+1)-8=0

The zeros of (2x+1)^2+5(2x+1)-8=0 are x=(-7+sqrt(57))/4 ,x=-(7+sqrt(57))/4

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(2 x + 1)^{2} + 5 (2 x + 1) - 8 = 0

x = \frac{- 7 + 57}{4}, x = - \frac{7 + 57}{4}

Solution steps

(2 x + 1)^{2} + 5 (2 x + 1) - 8 = 0

Expand

(2 x + 1)^{2} + 5 (2 x + 1) - 8 : 4 x^{2} + 14 x - 2

4 x^{2} + 14 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 4 ( - 2 )}{2 \cdot 4}

1 4^{2} - 4 \cdot 4 (- 2) = 257

x_{1, 2} = \frac{- 14 \pm 2 57}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 14 + 2 57}{2 \cdot 4}, x_{2} = \frac{- 14 - 2 57}{2 \cdot 4}

x = \frac{- 14 + 2 57}{2 \cdot 4} : \frac{- 7 + 57}{4}

x = \frac{- 14 - 2 57}{2 \cdot 4} : - \frac{7 + 57}{4}

The solutions to the quadratic equation are:

x = \frac{- 7 + 57}{4}, x = - \frac{7 + 57}{4}

x + 2 + \frac{x - 6}{2} = 10

\frac{2 + 3 x}{2} \geq 5

- 3 v + 12 v + 6 = 7 v + 30

2 x^{2} - 7 = - 7

x^{2} + (x + 1)^{2} = 80

What are the solutions to the equation (2x+1)^2+5(2x+1)-8=0 ?
The solutions to the equation (2x+1)^2+5(2x+1)-8=0 are x=(-7+sqrt(57))/4 ,x=-(7+sqrt(57))/4
Find the zeros of (2x+1)^2+5(2x+1)-8=0
The zeros of (2x+1)^2+5(2x+1)-8=0 are x=(-7+sqrt(57))/4 ,x=-(7+sqrt(57))/4