What are the solutions to the equation 2x^2+12x-80=0 ?

The solutions to the equation 2x^2+12x-80=0 are x=4,x=-10

Find the zeros of 2x^2+12x-80=0

The zeros of 2x^2+12x-80=0 are x=4,x=-10

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2 x^{2} + 12 x - 80 = 0

x = 4, x = - 10

Solution steps

2 x^{2} + 12 x - 80 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 2 ( - 80 )}{2 \cdot 2}

1 2^{2} - 4 \cdot 2 (- 80) = 28

x_{1, 2} = \frac{- 12 \pm 28}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 12 + 28}{2 \cdot 2}, x_{2} = \frac{- 12 - 28}{2 \cdot 2}

x = \frac{- 12 + 28}{2 \cdot 2} : 4

x = \frac{- 12 - 28}{2 \cdot 2} : - 10

The solutions to the quadratic equation are:

x = 4, x = - 10

3 x^{2} - 57 = 24 x - 9

x^{2} - 12 x + 35 = 3

3 z^{2} - 10 z + 2 = 2 z^{2}

12 x^{2} - 96 x + 135 = 0

What are the solutions to the equation 2x^2+12x-80=0 ?
The solutions to the equation 2x^2+12x-80=0 are x=4,x=-10
Find the zeros of 2x^2+12x-80=0
The zeros of 2x^2+12x-80=0 are x=4,x=-10