What are the solutions to the equation x^2+4x-120=0 ?

The solutions to the equation x^2+4x-120=0 are x=2(sqrt(31)-1),x=-2(1+sqrt(31))

Find the zeros of x^2+4x-120=0

The zeros of x^2+4x-120=0 are x=2(sqrt(31)-1),x=-2(1+sqrt(31))

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x^{2} + 4 x - 120 = 0

x = 2 (31 - 1), x = - 2 (1 + 31)

Decimal Notation

x = 9.13552 \dots, x = - 13.13552 \dots

Solution steps

x^{2} + 4 x - 120 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 120 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 120) = 431

x_{1, 2} = \frac{- 4 \pm 4 31}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 4 31}{2 \cdot 1}, x_{2} = \frac{- 4 - 4 31}{2 \cdot 1}

x = \frac{- 4 + 4 31}{2 \cdot 1} : 2 (31 - 1)

x = \frac{- 4 - 4 31}{2 \cdot 1} : - 2 (1 + 31)

The solutions to the quadratic equation are:

x = 2 (31 - 1), x = - 2 (1 + 31)

12 x^{2} + 8 = 20 x

3 t^{2} - 12 t - 36 = 0

(x) = 50000 + 200 x + 0.3 x^{2}

x (x - 1) = x + 3

9 x^{2} - 18 x + 3

What are the solutions to the equation x^2+4x-120=0 ?
The solutions to the equation x^2+4x-120=0 are x=2(sqrt(31)-1),x=-2(1+sqrt(31))
Find the zeros of x^2+4x-120=0
The zeros of x^2+4x-120=0 are x=2(sqrt(31)-1),x=-2(1+sqrt(31))