What are the solutions to the equation x^2+28x-128=0 ?

The solutions to the equation x^2+28x-128=0 are x=4,x=-32

Find the zeros of x^2+28x-128=0

The zeros of x^2+28x-128=0 are x=4,x=-32

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x^{2} + 28 x - 128 = 0

x = 4, x = - 32

Solution steps

x^{2} + 28 x - 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 \cdot 1 \cdot ( - 128 )}{2 \cdot 1}

2 8^{2} - 4 \cdot 1 \cdot (- 128) = 36

x_{1, 2} = \frac{- 28 \pm 36}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 28 + 36}{2 \cdot 1}, x_{2} = \frac{- 28 - 36}{2 \cdot 1}

x = \frac{- 28 + 36}{2 \cdot 1} : 4

x = \frac{- 28 - 36}{2 \cdot 1} : - 32

The solutions to the quadratic equation are:

x = 4, x = - 32

2 x^{2} + 5 x - 58 = 0

- 8 x^{2} + 3 x^{2} + 3 x - 1 = - 9 x - 8 - 6 x - 7 x^{2} + 7

a, 2 b y - 3 a y = 4 b^{2} - 9 a^{2}

(x - 5)^{2} - 10 = 0

5 (a - 2)^{2} + 3 (a + 1) = 25

What are the solutions to the equation x^2+28x-128=0 ?
The solutions to the equation x^2+28x-128=0 are x=4,x=-32
Find the zeros of x^2+28x-128=0
The zeros of x^2+28x-128=0 are x=4,x=-32