What are the solutions to the equation 12x-6x^2=-8 ?

The solutions to the equation 12x-6x^2=-8 are x=-(-3+sqrt(21))/3 ,x=(3+sqrt(21))/3

Find the zeros of 12x-6x^2=-8

The zeros of 12x-6x^2=-8 are x=-(-3+sqrt(21))/3 ,x=(3+sqrt(21))/3

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12 x - 6 x^{2} = - 8

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

Decimal Notation

x = - 0.52752 \dots, x = 2.52752 \dots

Solution steps

12 x - 6 x^{2} = - 8

Move

8

to the left side

12 x - 6 x^{2} + 8 = 0

Write in the standard form

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

- 6 x^{2} + 12 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 6 ) \cdot 8}{2 ( - 6 )}

1 2^{2} - 4 (- 6) \cdot 8 = 421

x_{1, 2} = \frac{- 12 \pm 4 21}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 12 + 4 21}{2 ( - 6 )}, x_{2} = \frac{- 12 - 4 21}{2 ( - 6 )}

x = \frac{- 12 + 4 21}{2 ( - 6 )} : - \frac{- 3 + 21}{3}

x = \frac{- 12 - 4 21}{2 ( - 6 )} : \frac{3 + 21}{3}

The solutions to the quadratic equation are:

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

x^{2} y + x y^{2} + 2 x yz + x y^{2} + y^{3} + 2 y^{2} z

\frac{y}{6} \leq 9

- \frac{1}{6} (x - 3) = \frac{1}{9} (x - \frac{3}{8})

lo g_{π} (1) = x

3 8^{2}

What are the solutions to the equation 12x-6x^2=-8 ?
The solutions to the equation 12x-6x^2=-8 are x=-(-3+sqrt(21))/3 ,x=(3+sqrt(21))/3
Find the zeros of 12x-6x^2=-8
The zeros of 12x-6x^2=-8 are x=-(-3+sqrt(21))/3 ,x=(3+sqrt(21))/3