What are the solutions to the equation 5x(4-x)=12 ?

The solutions to the equation 5x(4-x)=12 are x=(2(5-sqrt(10)))/5 ,x=(2(5+sqrt(10)))/5

Find the zeros of 5x(4-x)=12

The zeros of 5x(4-x)=12 are x=(2(5-sqrt(10)))/5 ,x=(2(5+sqrt(10)))/5

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5 x (4 - x) = 12

x = \frac{2 ( 5 - 10 )}{5}, x = \frac{2 ( 5 + 10 )}{5}

Decimal Notation

x = 0.73508 \dots, x = 3.26491 \dots

Solution steps

5 x (4 - x) = 12

Expand

5 x (4 - x) : 20 x - 5 x^{2}

20 x - 5 x^{2} = 12

Move

12

to the left side

20 x - 5 x^{2} - 12 = 0

Write in the standard form

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

- 5 x^{2} + 20 x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 5 ) ( - 12 )}{2 ( - 5 )}

2 0^{2} - 4 (- 5) (- 12) = 410

x_{1, 2} = \frac{- 20 \pm 4 10}{2 ( - 5 )}

Separate the solutions

x_{1} = \frac{- 20 + 4 10}{2 ( - 5 )}, x_{2} = \frac{- 20 - 4 10}{2 ( - 5 )}

x = \frac{- 20 + 4 10}{2 ( - 5 )} : \frac{2 ( 5 - 10 )}{5}

x = \frac{- 20 - 4 10}{2 ( - 5 )} : \frac{2 ( 5 + 10 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 5 - 10 )}{5}, x = \frac{2 ( 5 + 10 )}{5}

\frac{n ^{4} - 11 n ^{2} + 30}{n ^{4} - 7 n ^{2} + 10}

9 x^{2} - 76 x - 45

4 r^{2} + 2 r - 2 = 0

2 z^{2} + z = 6

What are the solutions to the equation 5x(4-x)=12 ?
The solutions to the equation 5x(4-x)=12 are x=(2(5-sqrt(10)))/5 ,x=(2(5+sqrt(10)))/5
Find the zeros of 5x(4-x)=12
The zeros of 5x(4-x)=12 are x=(2(5-sqrt(10)))/5 ,x=(2(5+sqrt(10)))/5