What are the solutions to the equation x^2+((4-x)/2)^2=16 ?

The solutions to the equation x^2+((4-x)/2)^2=16 are x=4,x=-12/5

Find the zeros of x^2+((4-x)/2)^2=16

The zeros of x^2+((4-x)/2)^2=16 are x=4,x=-12/5

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x^{2} + (\frac{4 - x}{2})^{2} = 16

x = 4, x = - \frac{12}{5}

Decimal Notation

x = 4, x = - 2.4

Solution steps

x^{2} + (\frac{4 - x}{2})^{2} = 16

Expand

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

4 - 2 x + \frac{5 x ^{2}}{4} = 16

Move

16

to the left side

\frac{5 x ^{2}}{4} - 2 x - 12 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot \frac{5}{4} (- 12) = 8

x_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot \frac{5}{4}}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 8}{2 \cdot \frac{5}{4}}, x_{2} = \frac{- ( - 2 ) - 8}{2 \cdot \frac{5}{4}}

x = \frac{- ( - 2 ) + 8}{2 \frac{5}{4}} : 4

x = \frac{- ( - 2 ) - 8}{2 \frac{5}{4}} : - \frac{12}{5}

The solutions to the quadratic equation are:

x = 4, x = - \frac{12}{5}

3 x^{2} = 81

m p^{2} + 2 m + n p^{2} = 2 n

- 3 x^{2} + 49 x = - 5 x + 183

3 x^{2} - 6 x - 10

m^{2} = 5 m + 14

What are the solutions to the equation x^2+((4-x)/2)^2=16 ?
The solutions to the equation x^2+((4-x)/2)^2=16 are x=4,x=-12/5
Find the zeros of x^2+((4-x)/2)^2=16
The zeros of x^2+((4-x)/2)^2=16 are x=4,x=-12/5