What are the solutions to the equation 3x^2+6x-62=7 ?

The solutions to the equation 3x^2+6x-62=7 are x=-1+2sqrt(6),x=-1-2sqrt(6)

Find the zeros of 3x^2+6x-62=7

The zeros of 3x^2+6x-62=7 are x=-1+2sqrt(6),x=-1-2sqrt(6)

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3 x^{2} + 6 x - 62 = 7

x = - 1 + 26, x = - 1 - 26

Decimal Notation

x = 3.89897 \dots, x = - 5.89897 \dots

Solution steps

3 x^{2} + 6 x - 62 = 7

Move

7

to the left side

3 x^{2} + 6 x - 69 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 69 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 69) = 126

x_{1, 2} = \frac{- 6 \pm 12 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 12 6}{2 \cdot 3}, x_{2} = \frac{- 6 - 12 6}{2 \cdot 3}

x = \frac{- 6 + 12 6}{2 \cdot 3} : - 1 + 26

x = \frac{- 6 - 12 6}{2 \cdot 3} : - 1 - 26

The solutions to the quadratic equation are:

x = - 1 + 26, x = - 1 - 26

t^{2} + t + 1 = (t - 1)^{2}

2 b^{3} + b^{5}

2 s \geq 6

29 x - 29 = 29 x - 11

- 3 = - 6 - (x + 15)

What are the solutions to the equation 3x^2+6x-62=7 ?
The solutions to the equation 3x^2+6x-62=7 are x=-1+2sqrt(6),x=-1-2sqrt(6)
Find the zeros of 3x^2+6x-62=7
The zeros of 3x^2+6x-62=7 are x=-1+2sqrt(6),x=-1-2sqrt(6)