What are the solutions to the equation 2x^2+28x+82=0 ?

The solutions to the equation 2x^2+28x+82=0 are x=-7+2sqrt(2),x=-7-2sqrt(2)

Find the zeros of 2x^2+28x+82=0

The zeros of 2x^2+28x+82=0 are x=-7+2sqrt(2),x=-7-2sqrt(2)

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2 x^{2} + 28 x + 82 = 0

x = - 7 + 22, x = - 7 - 22

Decimal Notation

x = - 4.17157 \dots, x = - 9.82842 \dots

Solution steps

2 x^{2} + 28 x + 82 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 2 \cdot 82}{2 \cdot 2}

2 8^{2} - 4 \cdot 2 \cdot 82 = 82

x_{1, 2} = \frac{- 28 \pm 8 2}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 28 + 8 2}{2 \cdot 2}, x_{2} = \frac{- 28 - 8 2}{2 \cdot 2}

x = \frac{- 28 + 8 2}{2 \cdot 2} : - 7 + 22

x = \frac{- 28 - 8 2}{2 \cdot 2} : - 7 - 22

The solutions to the quadratic equation are:

x = - 7 + 22, x = - 7 - 22

2 X^{2} + 8 = 0

2 x (x - 5) = (5 - 5)^{2}

x, 2 x^{2} + y^{3} = 0

a x^{2} = c

9 x^{2} - 6 x - 2 = 0

What are the solutions to the equation 2x^2+28x+82=0 ?
The solutions to the equation 2x^2+28x+82=0 are x=-7+2sqrt(2),x=-7-2sqrt(2)
Find the zeros of 2x^2+28x+82=0
The zeros of 2x^2+28x+82=0 are x=-7+2sqrt(2),x=-7-2sqrt(2)