What are the solutions to the equation 4x^2+20x+13=0 ?

The solutions to the equation 4x^2+20x+13=0 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2

Find the zeros of 4x^2+20x+13=0

The zeros of 4x^2+20x+13=0 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2

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4 x^{2} + 20 x + 13 = 0

x = \frac{- 5 + 2 3}{2}, x = - \frac{5 + 2 3}{2}

Decimal Notation

x = - 0.76794 \dots, x = - 4.23205 \dots

Solution steps

4 x^{2} + 20 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

2 0^{2} - 4 \cdot 4 \cdot 13 = 83

x_{1, 2} = \frac{- 20 \pm 8 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 20 + 8 3}{2 \cdot 4}, x_{2} = \frac{- 20 - 8 3}{2 \cdot 4}

x = \frac{- 20 + 8 3}{2 \cdot 4} : \frac{- 5 + 2 3}{2}

x = \frac{- 20 - 8 3}{2 \cdot 4} : - \frac{5 + 2 3}{2}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2 3}{2}, x = - \frac{5 + 2 3}{2}

2 x^{2} + x \geq 6

6 - x > 0

- 12 - 4

(\frac{3 m ^{5} r ^{3}}{4 p ^{8}})^{4}

3 (\frac{2}{3})^{2}

What are the solutions to the equation 4x^2+20x+13=0 ?
The solutions to the equation 4x^2+20x+13=0 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2
Find the zeros of 4x^2+20x+13=0
The zeros of 4x^2+20x+13=0 are x=(-5+2sqrt(3))/2 ,x=-(5+2sqrt(3))/2