What are the solutions to the equation 4x^2-8x-32=0 ?

The solutions to the equation 4x^2-8x-32=0 are x=4,x=-2

Find the zeros of 4x^2-8x-32=0

The zeros of 4x^2-8x-32=0 are x=4,x=-2

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4 x^{2} - 8 x - 32 = 0

x = 4, x = - 2

Solution steps

4 x^{2} - 8 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 4 ( - 32 )}{2 \cdot 4}

(- 8)^{2} - 4 \cdot 4 (- 32) = 24

x_{1, 2} = \frac{- ( - 8 ) \pm 24}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 24}{2 \cdot 4}, x_{2} = \frac{- ( - 8 ) - 24}{2 \cdot 4}

x = \frac{- ( - 8 ) + 24}{2 \cdot 4} : 4

x = \frac{- ( - 8 ) - 24}{2 \cdot 4} : - 2

The solutions to the quadratic equation are:

x = 4, x = - 2

10 m + 2 = 1 + \frac{m}{2}

- \frac{2}{3} + \frac{1}{3}

- 2 - q > 1

5 - \frac{1}{2} (x - 6) < 4

(\frac{π}{2}) + (\frac{π}{2})

What are the solutions to the equation 4x^2-8x-32=0 ?
The solutions to the equation 4x^2-8x-32=0 are x=4,x=-2
Find the zeros of 4x^2-8x-32=0
The zeros of 4x^2-8x-32=0 are x=4,x=-2