What are the solutions to the equation x^2-8x-128=0 ?

The solutions to the equation x^2-8x-128=0 are x=16,x=-8

Find the zeros of x^2-8x-128=0

The zeros of x^2-8x-128=0 are x=16,x=-8

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x^{2} - 8 x - 128 = 0

x = 16, x = - 8

Solution steps

x^{2} - 8 x - 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot ( - 128 )}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot (- 128) = 24

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

Separate the solutions

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

x = \frac{- ( - 8 ) + 24}{2 \cdot 1} : 16

x = \frac{- ( - 8 ) - 24}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

x = 16, x = - 8

\frac{25 - 3}{25}

\frac{7 π}{4} \cdot \frac{180}{π}

14

C + 35 < - 99

16 x^{2} + 0.5 + \frac{1}{256 x ^{2}}

What are the solutions to the equation x^2-8x-128=0 ?
The solutions to the equation x^2-8x-128=0 are x=16,x=-8
Find the zeros of x^2-8x-128=0
The zeros of x^2-8x-128=0 are x=16,x=-8