What are the solutions to the equation (2+3t)^2+(1-t)^2+(1+2t)^2=6 ?

The solutions to the equation (2+3t)^2+(1-t)^2+(1+2t)^2=6 are t=0,t=-1

Find the zeros of (2+3t)^2+(1-t)^2+(1+2t)^2=6

The zeros of (2+3t)^2+(1-t)^2+(1+2t)^2=6 are t=0,t=-1

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(2 + 3 t)^{2} + (1 - t)^{2} + (1 + 2 t)^{2} = 6

t = 0, t = - 1

Solution steps

(2 + 3 t)^{2} + (1 - t)^{2} + (1 + 2 t)^{2} = 6

Expand

(2 + 3 t)^{2} + (1 - t)^{2} + (1 + 2 t)^{2} : 14 t^{2} + 14 t + 6

14 t^{2} + 14 t + 6 = 6

Move

6

to the left side

14 t^{2} + 14 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 14 \cdot 0}{2 \cdot 14}

1 4^{2} - 4 \cdot 14 \cdot 0 = 14

t_{1, 2} = \frac{- 14 \pm 14}{2 \cdot 14}

Separate the solutions

t_{1} = \frac{- 14 + 14}{2 \cdot 14}, t_{2} = \frac{- 14 - 14}{2 \cdot 14}

t = \frac{- 14 + 14}{2 \cdot 14} : 0

t = \frac{- 14 - 14}{2 \cdot 14} : - 1

The solutions to the quadratic equation are:

t = 0, t = - 1

a, (a + b)^{2} = d

- r^{2} - 3 r - 2 = - 5 r^{2}

(x + 1)^{2} + (\frac{( 0 - 1 ) ^{2}}{2}) = 1

3 t^{2} - 8 t - 3 = 0

10 t^{2} + 33 t + 25 = 5

What are the solutions to the equation (2+3t)^2+(1-t)^2+(1+2t)^2=6 ?
The solutions to the equation (2+3t)^2+(1-t)^2+(1+2t)^2=6 are t=0,t=-1
Find the zeros of (2+3t)^2+(1-t)^2+(1+2t)^2=6
The zeros of (2+3t)^2+(1-t)^2+(1+2t)^2=6 are t=0,t=-1