What are the solutions to the equation (4-2t)^2+(-1+t)^2+(2-t)^2=9 ?

The solutions to the equation (4-2t)^2+(-1+t)^2+(2-t)^2=9 are t=3,t= 2/3

Find the zeros of (4-2t)^2+(-1+t)^2+(2-t)^2=9

The zeros of (4-2t)^2+(-1+t)^2+(2-t)^2=9 are t=3,t= 2/3

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(4 - 2 t)^{2} + (- 1 + t)^{2} + (2 - t)^{2} = 9

t = 3, t = \frac{2}{3}

Decimal Notation

t = 3, t = 0.66666 \dots

Solution steps

(4 - 2 t)^{2} + (- 1 + t)^{2} + (2 - t)^{2} = 9

Expand

(4 - 2 t)^{2} + (- 1 + t)^{2} + (2 - t)^{2} : 6 t^{2} - 22 t + 21

6 t^{2} - 22 t + 21 = 9

Move

9

to the left side

6 t^{2} - 22 t + 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 22 ) \pm ( - 22 ) ^{2} - 4 \cdot 6 \cdot 12}{2 \cdot 6}

(- 22)^{2} - 4 \cdot 6 \cdot 12 = 14

t_{1, 2} = \frac{- ( - 22 ) \pm 14}{2 \cdot 6}

Separate the solutions

t_{1} = \frac{- ( - 22 ) + 14}{2 \cdot 6}, t_{2} = \frac{- ( - 22 ) - 14}{2 \cdot 6}

t = \frac{- ( - 22 ) + 14}{2 \cdot 6} : 3

t = \frac{- ( - 22 ) - 14}{2 \cdot 6} : \frac{2}{3}

The solutions to the quadratic equation are:

t = 3, t = \frac{2}{3}

21 x^{2} + 21 x - 252 = 0

4 y^{2} = 28 y

x^{2} + 2 x - 195 = 0

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

12 x^{2} + 6 x + 11 = 0

What are the solutions to the equation (4-2t)^2+(-1+t)^2+(2-t)^2=9 ?
The solutions to the equation (4-2t)^2+(-1+t)^2+(2-t)^2=9 are t=3,t= 2/3
Find the zeros of (4-2t)^2+(-1+t)^2+(2-t)^2=9
The zeros of (4-2t)^2+(-1+t)^2+(2-t)^2=9 are t=3,t= 2/3