What are the solutions to the equation 4x^2+16x+13=0 ?

The solutions to the equation 4x^2+16x+13=0 are x=(-4+sqrt(3))/2 ,x=-(4+sqrt(3))/2

Find the zeros of 4x^2+16x+13=0

The zeros of 4x^2+16x+13=0 are x=(-4+sqrt(3))/2 ,x=-(4+sqrt(3))/2

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4 x^{2} + 16 x + 13 = 0

x = \frac{- 4 + 3}{2}, x = - \frac{4 + 3}{2}

Decimal Notation

x = - 1.13397 \dots, x = - 2.86602 \dots

Solution steps

4 x^{2} + 16 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 4 \cdot 13}{2 \cdot 4}

1 6^{2} - 4 \cdot 4 \cdot 13 = 43

x_{1, 2} = \frac{- 16 \pm 4 3}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- 16 + 4 3}{2 \cdot 4}, x_{2} = \frac{- 16 - 4 3}{2 \cdot 4}

x = \frac{- 16 + 4 3}{2 \cdot 4} : \frac{- 4 + 3}{2}

x = \frac{- 16 - 4 3}{2 \cdot 4} : - \frac{4 + 3}{2}

The solutions to the quadratic equation are:

x = \frac{- 4 + 3}{2}, x = - \frac{4 + 3}{2}

x^{2} - 20 x + 33 = 0

4 x^{2} - 20 = 5

4 y^{2} - 7 y - 4 = 0

117 = 0.0443213 (x)^{2}

0 = 30 - \frac{1}{2} \cdot 9.8 \cdot t^{2}

What are the solutions to the equation 4x^2+16x+13=0 ?
The solutions to the equation 4x^2+16x+13=0 are x=(-4+sqrt(3))/2 ,x=-(4+sqrt(3))/2
Find the zeros of 4x^2+16x+13=0
The zeros of 4x^2+16x+13=0 are x=(-4+sqrt(3))/2 ,x=-(4+sqrt(3))/2