What are the solutions to the equation 2x(2x+1)=5 ?

The solutions to the equation 2x(2x+1)=5 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4

Find the zeros of 2x(2x+1)=5

The zeros of 2x(2x+1)=5 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4

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2 x (2 x + 1) = 5

x = \frac{- 1 + 21}{4}, x = - \frac{1 + 21}{4}

Decimal Notation

x = 0.89564 \dots, x = - 1.39564 \dots

Solution steps

2 x (2 x + 1) = 5

Expand

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

4 x^{2} + 2 x = 5

Move

5

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 4 (- 5) = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 \cdot 4}, x_{2} = \frac{- 2 - 2 21}{2 \cdot 4}

x = \frac{- 2 + 2 21}{2 \cdot 4} : \frac{- 1 + 21}{4}

x = \frac{- 2 - 2 21}{2 \cdot 4} : - \frac{1 + 21}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{4}, x = - \frac{1 + 21}{4}

x^{2} + \frac{3}{3} x + \frac{1}{3} = 0

x^{2} + 6 x + 9 = 1

4 a x^{2} - 4 x - (a - 2) = 0

6 x^{2} + 42 x + 60 = 0

x, f = m x^{2}

What are the solutions to the equation 2x(2x+1)=5 ?
The solutions to the equation 2x(2x+1)=5 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4
Find the zeros of 2x(2x+1)=5
The zeros of 2x(2x+1)=5 are x=(-1+sqrt(21))/4 ,x=-(1+sqrt(21))/4