What are the solutions to the equation (4x+1)(4x+3)=5 ?

The solutions to the equation (4x+1)(4x+3)=5 are x=(-2+sqrt(6))/4 ,x=-(2+sqrt(6))/4

Find the zeros of (4x+1)(4x+3)=5

The zeros of (4x+1)(4x+3)=5 are x=(-2+sqrt(6))/4 ,x=-(2+sqrt(6))/4

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

x = \frac{- 2 + 6}{4}, x = - \frac{2 + 6}{4}

Decimal Notation

x = 0.11237 \dots, x = - 1.11237 \dots

Solution steps

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

Expand

(4 x + 1) (4 x + 3) : 16 x^{2} + 16 x + 3

16 x^{2} + 16 x + 3 = 5

Move

5

to the left side

16 x^{2} + 16 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 16 ( - 2 )}{2 \cdot 16}

1 6^{2} - 4 \cdot 16 (- 2) = 86

x_{1, 2} = \frac{- 16 \pm 8 6}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- 16 + 8 6}{2 \cdot 16}, x_{2} = \frac{- 16 - 8 6}{2 \cdot 16}

x = \frac{- 16 + 8 6}{2 \cdot 16} : \frac{- 2 + 6}{4}

x = \frac{- 16 - 8 6}{2 \cdot 16} : - \frac{2 + 6}{4}

The solutions to the quadratic equation are:

x = \frac{- 2 + 6}{4}, x = - \frac{2 + 6}{4}

(1 - x) ln (π) = x

1 0^{8}

- 4 b^{4} - 500 b

a^{2} + 4 a + 5 = 0

What are the solutions to the equation (4x+1)(4x+3)=5 ?
The solutions to the equation (4x+1)(4x+3)=5 are x=(-2+sqrt(6))/4 ,x=-(2+sqrt(6))/4
Find the zeros of (4x+1)(4x+3)=5
The zeros of (4x+1)(4x+3)=5 are x=(-2+sqrt(6))/4 ,x=-(2+sqrt(6))/4