What are the solutions to the equation solvefor x,x^2+12x-28=0 ?

The solutions to the equation solvefor x,x^2+12x-28=0 are x=2,x=-14

Find the zeros of solvefor x,x^2+12x-28=0

The zeros of solvefor x,x^2+12x-28=0 are x=2,x=-14

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x, x^{2} + 12 x - 28 = 0

x = 2, x = - 14

Solution steps

x^{2} + 12 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 28 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 28) = 16

x_{1, 2} = \frac{- 12 \pm 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 16}{2 \cdot 1}, x_{2} = \frac{- 12 - 16}{2 \cdot 1}

x = \frac{- 12 + 16}{2 \cdot 1} : 2

x = \frac{- 12 - 16}{2 \cdot 1} : - 14

The solutions to the quadratic equation are:

x = 2, x = - 14

x^{2} - 12 x + 46 = 0

75 = - \frac{1}{4} x^{2} + 5 x

2 w^{2} - w - 21 = 0

n^{2} - 6 n + 7 = 0

(300 + 10 x) (1.6 - 0.01 x)

What are the solutions to the equation solvefor x,x^2+12x-28=0 ?
The solutions to the equation solvefor x,x^2+12x-28=0 are x=2,x=-14
Find the zeros of solvefor x,x^2+12x-28=0
The zeros of solvefor x,x^2+12x-28=0 are x=2,x=-14