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

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

Find the zeros of X^2+8X-128=0

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

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X^{2} + 8 X - 128 = 0

X = 8, X = - 16

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} : 8

X = \frac{- 8 - 24}{2 \cdot 1} : - 16

The solutions to the quadratic equation are:

X = 8, X = - 16

x^{3} + x^{2} + 9 x + 9

\frac{17}{20} = t + (- \frac{13}{20})

2 x - 5 > 9

\frac{3}{4} \cdot \frac{7}{6}

u^{3} - 8

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