What are the solutions to the equation 2x^2+24x+32=0 ?

The solutions to the equation 2x^2+24x+32=0 are x=2(sqrt(5)-3),x=-2(3+sqrt(5))

Find the zeros of 2x^2+24x+32=0

The zeros of 2x^2+24x+32=0 are x=2(sqrt(5)-3),x=-2(3+sqrt(5))

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2 x^{2} + 24 x + 32 = 0

x = 2 (5 - 3), x = - 2 (3 + 5)

Decimal Notation

x = - 1.52786 \dots, x = - 10.47213 \dots

Solution steps

2 x^{2} + 24 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 2 \cdot 32}{2 \cdot 2}

2 4^{2} - 4 \cdot 2 \cdot 32 = 85

x_{1, 2} = \frac{- 24 \pm 8 5}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 24 + 8 5}{2 \cdot 2}, x_{2} = \frac{- 24 - 8 5}{2 \cdot 2}

x = \frac{- 24 + 8 5}{2 \cdot 2} : 2 (5 - 3)

x = \frac{- 24 - 8 5}{2 \cdot 2} : - 2 (3 + 5)

The solutions to the quadratic equation are:

x = 2 (5 - 3), x = - 2 (3 + 5)

x^{2} + 15 x + 54 = 0

x^{2} = 6 x + 1

r^{2} - 9

x^{2} - 2 x + 1 = 144

x^{2} + 0.48 x - 0.0276 = 0

What are the solutions to the equation 2x^2+24x+32=0 ?
The solutions to the equation 2x^2+24x+32=0 are x=2(sqrt(5)-3),x=-2(3+sqrt(5))
Find the zeros of 2x^2+24x+32=0
The zeros of 2x^2+24x+32=0 are x=2(sqrt(5)-3),x=-2(3+sqrt(5))