What are the solutions to the equation 3x^2+6x-29=-4 ?

The solutions to the equation 3x^2+6x-29=-4 are x=(-3+2sqrt(21))/3 ,x=-(3+2sqrt(21))/3

Find the zeros of 3x^2+6x-29=-4

The zeros of 3x^2+6x-29=-4 are x=(-3+2sqrt(21))/3 ,x=-(3+2sqrt(21))/3

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3 x^{2} + 6 x - 29 = - 4

x = \frac{- 3 + 2 21}{3}, x = - \frac{3 + 2 21}{3}

Decimal Notation

x = 2.05505 \dots, x = - 4.05505 \dots

Solution steps

3 x^{2} + 6 x - 29 = - 4

Move

4

to the left side

3 x^{2} + 6 x - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 ( - 25 )}{2 \cdot 3}

6^{2} - 4 \cdot 3 (- 25) = 421

x_{1, 2} = \frac{- 6 \pm 4 21}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 4 21}{2 \cdot 3}, x_{2} = \frac{- 6 - 4 21}{2 \cdot 3}

x = \frac{- 6 + 4 21}{2 \cdot 3} : \frac{- 3 + 2 21}{3}

x = \frac{- 6 - 4 21}{2 \cdot 3} : - \frac{3 + 2 21}{3}

The solutions to the quadratic equation are:

x = \frac{- 3 + 2 21}{3}, x = - \frac{3 + 2 21}{3}

y, x^{2} y = y - 7

2 x^{2} - 7 x + 4

r^{2} - 6 r + 12 = 0

(1 - k) x^{2} + x + k = 0

81 = \frac{n}{2} (1.5 + n)

What are the solutions to the equation 3x^2+6x-29=-4 ?
The solutions to the equation 3x^2+6x-29=-4 are x=(-3+2sqrt(21))/3 ,x=-(3+2sqrt(21))/3
Find the zeros of 3x^2+6x-29=-4
The zeros of 3x^2+6x-29=-4 are x=(-3+2sqrt(21))/3 ,x=-(3+2sqrt(21))/3