What are the solutions to the equation 2x^2-12x-32=0 ?

The solutions to the equation 2x^2-12x-32=0 are x=8,x=-2

Find the zeros of 2x^2-12x-32=0

The zeros of 2x^2-12x-32=0 are x=8,x=-2

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2 x^{2} - 12 x - 32 = 0

x = 8, x = - 2

Solution steps

2 x^{2} - 12 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 2 ( - 32 )}{2 \cdot 2}

(- 12)^{2} - 4 \cdot 2 (- 32) = 20

x_{1, 2} = \frac{- ( - 12 ) \pm 20}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 20}{2 \cdot 2}, x_{2} = \frac{- ( - 12 ) - 20}{2 \cdot 2}

x = \frac{- ( - 12 ) + 20}{2 \cdot 2} : 8

x = \frac{- ( - 12 ) - 20}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = 8, x = - 2

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- 4 y^{4} + 6 y^{2} - 14 y

2 (4 x + 1) (x - 1) + 3 = 0

What are the solutions to the equation 2x^2-12x-32=0 ?
The solutions to the equation 2x^2-12x-32=0 are x=8,x=-2
Find the zeros of 2x^2-12x-32=0
The zeros of 2x^2-12x-32=0 are x=8,x=-2