What are the solutions to the equation x^2+12x-48=0 ?

The solutions to the equation x^2+12x-48=0 are x=2(sqrt(21)-3),x=-2(3+sqrt(21))

Find the zeros of x^2+12x-48=0

The zeros of x^2+12x-48=0 are x=2(sqrt(21)-3),x=-2(3+sqrt(21))

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x^{2} + 12 x - 48 = 0

x = 2 (21 - 3), x = - 2 (3 + 21)

Decimal Notation

x = 3.16515 \dots, x = - 15.16515 \dots

Solution steps

x^{2} + 12 x - 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 48 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 48) = 421

x_{1, 2} = \frac{- 12 \pm 4 21}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4 21}{2 \cdot 1}, x_{2} = \frac{- 12 - 4 21}{2 \cdot 1}

x = \frac{- 12 + 4 21}{2 \cdot 1} : 2 (21 - 3)

x = \frac{- 12 - 4 21}{2 \cdot 1} : - 2 (3 + 21)

The solutions to the quadratic equation are:

x = 2 (21 - 3), x = - 2 (3 + 21)

3 m^{2} - 17 m - 28 = 0

5 r^{2} - 44 r + 120 = - 30 + 11 r

x^{2} - 10 x + 3

0 = x^{2} + 8 x - 20

- 8 x^{2} + 80 = 0

What are the solutions to the equation x^2+12x-48=0 ?
The solutions to the equation x^2+12x-48=0 are x=2(sqrt(21)-3),x=-2(3+sqrt(21))
Find the zeros of x^2+12x-48=0
The zeros of x^2+12x-48=0 are x=2(sqrt(21)-3),x=-2(3+sqrt(21))