What are the solutions to the equation 3x^2+42x+132=0 ?

The solutions to the equation 3x^2+42x+132=0 are x=-7+sqrt(5),x=-7-sqrt(5)

Find the zeros of 3x^2+42x+132=0

The zeros of 3x^2+42x+132=0 are x=-7+sqrt(5),x=-7-sqrt(5)

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3 x^{2} + 42 x + 132 = 0

x = - 7 + 5, x = - 7 - 5

Decimal Notation

x = - 4.76393 \dots, x = - 9.23606 \dots

Solution steps

3 x^{2} + 42 x + 132 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 42 \pm 4 2 ^{2} - 4 \cdot 3 \cdot 132}{2 \cdot 3}

4 2^{2} - 4 \cdot 3 \cdot 132 = 65

x_{1, 2} = \frac{- 42 \pm 6 5}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 42 + 6 5}{2 \cdot 3}, x_{2} = \frac{- 42 - 6 5}{2 \cdot 3}

x = \frac{- 42 + 6 5}{2 \cdot 3} : - 7 + 5

x = \frac{- 42 - 6 5}{2 \cdot 3} : - 7 - 5

The solutions to the quadratic equation are:

x = - 7 + 5, x = - 7 - 5

x, x^{2} + 2 x - 3 = 0

(x + 1) (x + 2) = 6

10 x^{2} - 250 = 0

x^{2} + 5 = 9

x^{2} + 2 = 6

What are the solutions to the equation 3x^2+42x+132=0 ?
The solutions to the equation 3x^2+42x+132=0 are x=-7+sqrt(5),x=-7-sqrt(5)
Find the zeros of 3x^2+42x+132=0
The zeros of 3x^2+42x+132=0 are x=-7+sqrt(5),x=-7-sqrt(5)