What are the solutions to the equation x(22-x)=112 ?

The solutions to the equation x(22-x)=112 are x=8,x=14

Find the zeros of x(22-x)=112

The zeros of x(22-x)=112 are x=8,x=14

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x (22 - x) = 112

x = 8, x = 14

Solution steps

x (22 - x) = 112

Expand

x (22 - x) : 22 x - x^{2}

22 x - x^{2} = 112

Move

112

to the left side

22 x - x^{2} - 112 = 0

Write in the standard form

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

- x^{2} + 22 x - 112 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 22 \pm 2 2 ^{2} - 4 ( - 1 ) ( - 112 )}{2 ( - 1 )}

2 2^{2} - 4 (- 1) (- 112) = 6

x_{1, 2} = \frac{- 22 \pm 6}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 22 + 6}{2 ( - 1 )}, x_{2} = \frac{- 22 - 6}{2 ( - 1 )}

x = \frac{- 22 + 6}{2 ( - 1 )} : 8

x = \frac{- 22 - 6}{2 ( - 1 )} : 14

The solutions to the quadratic equation are:

x = 8, x = 14

\frac{2}{3} x^{2} - 10 = - 28

8 x^{2} - 3 x - 7 = 0

3 k^{2} + 5 k + 1 = 0

2 x^{2} - 4 x = - 3

z^{2} - 9 = 0

What are the solutions to the equation x(22-x)=112 ?
The solutions to the equation x(22-x)=112 are x=8,x=14
Find the zeros of x(22-x)=112
The zeros of x(22-x)=112 are x=8,x=14