What are the solutions to the equation A^2=2x(4x+1) ?

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

Find the zeros of A^2=2x(4x+1)

The zeros of A^2=2x(4x+1) are x=(-1+sqrt(8A^2+1))/8 ,x=-(sqrt(8A^2+1)+1)/8

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A^{2} = 2 x (4 x + 1)

x = \frac{- 1 + 8 A ^{2} + 1}{8}, x = - \frac{8 A ^{2} + 1 + 1}{8}

Solution steps

A^{2} = 2 x (4 x + 1)

Expand

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

A^{2} = 8 x^{2} + 2 x

Switch sides

8 x^{2} + 2 x = A^{2}

Move

A^{2}

to the left side

8 x^{2} + 2 x - A^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 8 ( - A ^{2} )}{2 \cdot 8}

Simplify

2^{2} - 4 \cdot 8 (- A^{2}) : 2 1 + 8 A^{2}

x_{1, 2} = \frac{- 2 \pm 2 1 + 8 A ^{2}}{2 \cdot 8}

Separate the solutions

x_{1} = \frac{- 2 + 2 1 + 8 A ^{2}}{2 \cdot 8}, x_{2} = \frac{- 2 - 2 1 + 8 A ^{2}}{2 \cdot 8}

x = \frac{- 2 + 2 1 + 8 A ^{2}}{2 \cdot 8} : \frac{- 1 + 8 A ^{2} + 1}{8}

x = \frac{- 2 - 2 1 + 8 A ^{2}}{2 \cdot 8} : - \frac{8 A ^{2} + 1 + 1}{8}

The solutions to the quadratic equation are:

x = \frac{- 1 + 8 A ^{2} + 1}{8}, x = - \frac{8 A ^{2} + 1 + 1}{8}

x^{2} + 30 x + c

x^{2} - 8 x = - 2

(x - 150)^{2} + 16

2 x^{2} - x = 120

x^{2} - 9 x - 20 = - 2 x - 2

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