What are the solutions to the equation 3x(x+4)+6(x^2-1)=-10 ?

The solutions to the equation 3x(x+4)+6(x^2-1)=-10 are x=-2/3

Find the zeros of 3x(x+4)+6(x^2-1)=-10

The zeros of 3x(x+4)+6(x^2-1)=-10 are x=-2/3

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3 x (x + 4) + 6 (x^{2} - 1) = - 10

x = - \frac{2}{3}

Decimal Notation

x = - 0.66666 \dots

Solution steps

3 x (x + 4) + 6 (x^{2} - 1) = - 10

Expand

3 x (x + 4) + 6 (x^{2} - 1) : 9 x^{2} + 12 x - 6

9 x^{2} + 12 x - 6 = - 10

Move

10

to the left side

9 x^{2} + 12 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 9 \cdot 4}{2 \cdot 9}

1 2^{2} - 4 \cdot 9 \cdot 4 = 0

x_{1, 2} = \frac{- 12 \pm 0}{2 \cdot 9}

x = \frac{- 12}{2 \cdot 9}

\frac{- 12}{2 \cdot 9} = - \frac{2}{3}

x = - \frac{2}{3}

The solution to the quadratic equation is:

x = - \frac{2}{3}

c, m \cdot b \cdot a = n \cdot k \cdot c^{2} \cdot s \cdot p + m \cdot c \cdot a

x + 2 = - 3 x^{2}

x^{2} + 14 x + 65 = - 2

x, x^{2} + y^{2} + 4 x - 2 3 y + 3 = 0

2 x^{2} + x + 15 = 0

What are the solutions to the equation 3x(x+4)+6(x^2-1)=-10 ?
The solutions to the equation 3x(x+4)+6(x^2-1)=-10 are x=-2/3
Find the zeros of 3x(x+4)+6(x^2-1)=-10
The zeros of 3x(x+4)+6(x^2-1)=-10 are x=-2/3