What are the solutions to the equation x^2+12x-32=0 ?

The solutions to the equation x^2+12x-32=0 are x=2(sqrt(17)-3),x=-2(3+sqrt(17))

Find the zeros of x^2+12x-32=0

The zeros of x^2+12x-32=0 are x=2(sqrt(17)-3),x=-2(3+sqrt(17))

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x^{2} + 12 x - 32 = 0

x = 2 (17 - 3), x = - 2 (3 + 17)

Decimal Notation

x = 2.24621 \dots, x = - 14.24621 \dots

Solution steps

x^{2} + 12 x - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 32 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 32) = 417

x_{1, 2} = \frac{- 12 \pm 4 17}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4 17}{2 \cdot 1}, x_{2} = \frac{- 12 - 4 17}{2 \cdot 1}

x = \frac{- 12 + 4 17}{2 \cdot 1} : 2 (17 - 3)

x = \frac{- 12 - 4 17}{2 \cdot 1} : - 2 (3 + 17)

The solutions to the quadratic equation are:

x = 2 (17 - 3), x = - 2 (3 + 17)

3 y^{2} + 3 y - 45

24 - 8

(x - 1) (x - 2) (x - 3) \leq 0

3 x - 4 \leq 8 x + 6 \leq 14

2 x - 6 y + 8

What are the solutions to the equation x^2+12x-32=0 ?
The solutions to the equation x^2+12x-32=0 are x=2(sqrt(17)-3),x=-2(3+sqrt(17))
Find the zeros of x^2+12x-32=0
The zeros of x^2+12x-32=0 are x=2(sqrt(17)-3),x=-2(3+sqrt(17))