What are the solutions to the equation 10/3 =10x-5x^2 ?

The solutions to the equation 10/3 =10x-5x^2 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

Find the zeros of 10/3 =10x-5x^2

The zeros of 10/3 =10x-5x^2 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

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\frac{10}{3} = 10 x - 5 x^{2}

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

Decimal Notation

x = 0.42264 \dots, x = 1.57735 \dots

Solution steps

\frac{10}{3} = 10 x - 5 x^{2}

Multiply both sides by

3

10 = 30 x - 15 x^{2}

Switch sides

30 x - 15 x^{2} = 10

Move

10

to the left side

30 x - 15 x^{2} - 10 = 0

Write in the standard form

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

- 15 x^{2} + 30 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 ( - 15 ) ( - 10 )}{2 ( - 15 )}

3 0^{2} - 4 (- 15) (- 10) = 103

x_{1, 2} = \frac{- 30 \pm 10 3}{2 ( - 15 )}

Separate the solutions

x_{1} = \frac{- 30 + 10 3}{2 ( - 15 )}, x_{2} = \frac{- 30 - 10 3}{2 ( - 15 )}

x = \frac{- 30 + 10 3}{2 ( - 15 )} : \frac{3 - 3}{3}

x = \frac{- 30 - 10 3}{2 ( - 15 )} : \frac{3 + 3}{3}

The solutions to the quadratic equation are:

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

\frac{2}{3} x - 4 > 9

9^{l o g_{3} (5)} = x

\frac{2 ^{2}}{2 ^{- 9}}

\frac{r}{4} = 250

2 x - 1 \geq 1

What are the solutions to the equation 10/3 =10x-5x^2 ?
The solutions to the equation 10/3 =10x-5x^2 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3
Find the zeros of 10/3 =10x-5x^2
The zeros of 10/3 =10x-5x^2 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3