What are the solutions to the equation -8x+10-3x^2=-12x ?

The solutions to the equation -8x+10-3x^2=-12x are x=-(-2+sqrt(34))/3 ,x=(2+sqrt(34))/3

Find the zeros of -8x+10-3x^2=-12x

The zeros of -8x+10-3x^2=-12x are x=-(-2+sqrt(34))/3 ,x=(2+sqrt(34))/3

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- 8 x + 10 - 3 x^{2} = - 12 x

x = - \frac{- 2 + 34}{3}, x = \frac{2 + 34}{3}

Decimal Notation

x = - 1.27698 \dots, x = 2.61031 \dots

Solution steps

- 8 x + 10 - 3 x^{2} = - 12 x

Move

12 x

to the left side

- 3 x^{2} + 4 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 ( - 3 ) \cdot 10}{2 ( - 3 )}

4^{2} - 4 (- 3) \cdot 10 = 234

x_{1, 2} = \frac{- 4 \pm 2 34}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 4 + 2 34}{2 ( - 3 )}, x_{2} = \frac{- 4 - 2 34}{2 ( - 3 )}

x = \frac{- 4 + 2 34}{2 ( - 3 )} : - \frac{- 2 + 34}{3}

x = \frac{- 4 - 2 34}{2 ( - 3 )} : \frac{2 + 34}{3}

The solutions to the quadratic equation are:

x = - \frac{- 2 + 34}{3}, x = \frac{2 + 34}{3}

3 + 5 x - 2 \leq 6 x - 7 - 9 x

2 - 7 (4 - x) = 9

\frac{1}{\frac{1}{15} + \frac{1}{10}}

\frac{x - 3}{x ^{2} - 49} < 0

\frac{π}{2} - 2

What are the solutions to the equation -8x+10-3x^2=-12x ?
The solutions to the equation -8x+10-3x^2=-12x are x=-(-2+sqrt(34))/3 ,x=(2+sqrt(34))/3
Find the zeros of -8x+10-3x^2=-12x
The zeros of -8x+10-3x^2=-12x are x=-(-2+sqrt(34))/3 ,x=(2+sqrt(34))/3