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 ( - 2 ) \cdot 32}{2 ( - 2 )}

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 8, x = 2

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

y, y^{2} = C x^{4} + 3 x

x, x^{2} - 3 x y^{3} = 10

(2 x - 3)^{2} - 118 + (x - 6)^{2} = (x - 5)^{2}

6 = x^{2} - 10 x

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