What are the solutions to the equation 2x^2+24x+8=0 ?

The solutions to the equation 2x^2+24x+8=0 are x=2(2sqrt(2)-3),x=-2(3+2sqrt(2))

Find the zeros of 2x^2+24x+8=0

The zeros of 2x^2+24x+8=0 are x=2(2sqrt(2)-3),x=-2(3+2sqrt(2))

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2 x^{2} + 24 x + 8 = 0

x = 2 (22 - 3), x = - 2 (3 + 22)

Decimal Notation

x = - 0.34314 \dots, x = - 11.65685 \dots

Solution steps

2 x^{2} + 24 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

2 4^{2} - 4 \cdot 2 \cdot 8 = 162

x_{1, 2} = \frac{- 24 \pm 16 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 24 + 16 2}{2 \cdot 2}, x_{2} = \frac{- 24 - 16 2}{2 \cdot 2}

x = \frac{- 24 + 16 2}{2 \cdot 2} : 2 (22 - 3)

x = \frac{- 24 - 16 2}{2 \cdot 2} : - 2 (3 + 22)

The solutions to the quadratic equation are:

x = 2 (22 - 3), x = - 2 (3 + 22)

(3 x + 6) (6 x + 5) = 0

(3 x - 1) (2 x + 1) = 0

(x + 3)^{2} = 10 (3 x - 11)

t, s = - 16 t^{2} + 100 t

- x^{2} + 6 x - 16 = 0

What are the solutions to the equation 2x^2+24x+8=0 ?
The solutions to the equation 2x^2+24x+8=0 are x=2(2sqrt(2)-3),x=-2(3+2sqrt(2))
Find the zeros of 2x^2+24x+8=0
The zeros of 2x^2+24x+8=0 are x=2(2sqrt(2)-3),x=-2(3+2sqrt(2))