What are the solutions to the equation x^2+12x-4=0 ?

The solutions to the equation x^2+12x-4=0 are x=2(sqrt(10)-3),x=-2(3+sqrt(10))

Find the zeros of x^2+12x-4=0

The zeros of x^2+12x-4=0 are x=2(sqrt(10)-3),x=-2(3+sqrt(10))

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x^{2} + 12 x - 4 = 0

x = 2 (10 - 3), x = - 2 (3 + 10)

Solution steps

x^{2} + 12 x - 4 = 0

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 1 \cdot (- 4) = 410

x_{1, 2} = \frac{- 12 \pm 4 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4 10}{2 \cdot 1}, x_{2} = \frac{- 12 - 4 10}{2 \cdot 1}

x = \frac{- 12 + 4 10}{2 \cdot 1} : 2 (10 - 3)

x = \frac{- 12 - 4 10}{2 \cdot 1} : - 2 (3 + 10)

The solutions to the quadratic equation are:

x = 2 (10 - 3), x = - 2 (3 + 10)

2 v^{2} (v + 5) - 7 (v + 5)

4 - 3 a = 7 - 2 (2 a + 6)

\frac{3 x}{4} < x - \frac{x + 2}{2}

- 5 b + 2 = 17

2 < ∣ x - 6 ∣ < 4

What are the solutions to the equation x^2+12x-4=0 ?
The solutions to the equation x^2+12x-4=0 are x=2(sqrt(10)-3),x=-2(3+sqrt(10))
Find the zeros of x^2+12x-4=0
The zeros of x^2+12x-4=0 are x=2(sqrt(10)-3),x=-2(3+sqrt(10))