What are the solutions to the equation x(4-x)+5x=x^2+3x ?

The solutions to the equation x(4-x)+5x=x^2+3x are x=0,x=3

Find the zeros of x(4-x)+5x=x^2+3x

The zeros of x(4-x)+5x=x^2+3x are x=0,x=3

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x (4 - x) + 5 x = x^{2} + 3 x

x = 0, x = 3

Solution steps

x (4 - x) + 5 x = x^{2} + 3 x

Expand

x (4 - x) + 5 x : - x^{2} + 9 x

- x^{2} + 9 x = x^{2} + 3 x

Move

3 x

to the left side

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

Move

x^{2}

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 2 ) \cdot 0}{2 ( - 2 )}

6^{2} - 4 (- 2) \cdot 0 = 6

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

Separate the solutions

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

x = \frac{- 6 + 6}{2 ( - 2 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 3

6 x^{2} - 29 x + 35 = 0

14.6689 = x^{2} + x

(x - 2)^{3} - x^{3} + 6 x + 7 = 0

x, x^{2} + y^{2} + z^{2} - 2 x + 4 y + 5 = 0

15 x^{2} + 15 = 34 x

What are the solutions to the equation x(4-x)+5x=x^2+3x ?
The solutions to the equation x(4-x)+5x=x^2+3x are x=0,x=3
Find the zeros of x(4-x)+5x=x^2+3x
The zeros of x(4-x)+5x=x^2+3x are x=0,x=3