What are the solutions to the equation 2x(3-x)=3 ?

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

Find the zeros of 2x(3-x)=3

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

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2 x (3 - x) = 3

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

Decimal Notation

x = 0.63397 \dots, x = 2.36602 \dots

Solution steps

2 x (3 - x) = 3

Expand

2 x (3 - x) : 6 x - 2 x^{2}

6 x - 2 x^{2} = 3

Move

3

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 2 ) ( - 3 )}{2 ( - 2 )}

6^{2} - 4 (- 2) (- 3) = 23

x_{1, 2} = \frac{- 6 \pm 2 3}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 6 + 2 3}{2 ( - 2 )}, x_{2} = \frac{- 6 - 2 3}{2 ( - 2 )}

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

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

The solutions to the quadratic equation are:

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

- x^{2} - 7 x = 0

- 2 x (x - 1) = x^{2} - 5 x

x^{2} + 96 = 0

x^{2} - 1 = x + 1

x = x^{2} - 4

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