What are the solutions to the equation 9x^2-4=-12x ?

The solutions to the equation 9x^2-4=-12x are x=(2(sqrt(2)-1))/3 ,x=-(2(1+sqrt(2)))/3

Find the zeros of 9x^2-4=-12x

The zeros of 9x^2-4=-12x are x=(2(sqrt(2)-1))/3 ,x=-(2(1+sqrt(2)))/3

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9 x^{2} - 4 = - 12 x

x = \frac{2 ( 2 - 1 )}{3}, x = - \frac{2 ( 1 + 2 )}{3}

Decimal Notation

x = 0.27614 \dots, x = - 1.60947 \dots

Solution steps

9 x^{2} - 4 = - 12 x

Move

12 x

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1 2^{2} - 4 \cdot 9 (- 4) = 122

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

Separate the solutions

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

x = \frac{- 12 + 12 2}{2 \cdot 9} : \frac{2 ( 2 - 1 )}{3}

x = \frac{- 12 - 12 2}{2 \cdot 9} : - \frac{2 ( 1 + 2 )}{3}

The solutions to the quadratic equation are:

x = \frac{2 ( 2 - 1 )}{3}, x = - \frac{2 ( 1 + 2 )}{3}

x, 5 x^{2} + 4 x y + y^{2} = 1

\frac{1}{2} x^{2} + 2 x - 1 = 5

6 x^{2} - 12 x + 3 = 0

y^{2} = 13 y - 42

x^{2} + 20 x - 96 = 0

What are the solutions to the equation 9x^2-4=-12x ?
The solutions to the equation 9x^2-4=-12x are x=(2(sqrt(2)-1))/3 ,x=-(2(1+sqrt(2)))/3
Find the zeros of 9x^2-4=-12x
The zeros of 9x^2-4=-12x are x=(2(sqrt(2)-1))/3 ,x=-(2(1+sqrt(2)))/3