What are the solutions to the equation t(t)=25-2t,0<= t<= 10 ?

The solutions to the equation t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)

Find the zeros of t(t)=25-2t,0<= t<= 10

The zeros of t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)

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The solutions to the equation t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)

The zeros of t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)

t (t) = 25 - 2 t, 0 \leq t \leq 10

t = - 1 + 26

Decimal Notation

t = 4.09901 \dots

Solution steps

t (t) = 25 - 2 t

Expand

t (t) : t^{2}

t^{2} = 25 - 2 t

Move

2 t

to the left side

t^{2} + 2 t = 25

Move

25

to the left side

t^{2} + 2 t - 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 25) = 226

t_{1, 2} = \frac{- 2 \pm 2 26}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 2 26}{2 \cdot 1}, t_{2} = \frac{- 2 - 2 26}{2 \cdot 1}

t = \frac{- 2 + 2 26}{2 \cdot 1} : - 1 + 26

t = \frac{- 2 - 2 26}{2 \cdot 1} : - 1 - 26

The solutions to the quadratic equation are:

t = - 1 + 26, t = - 1 - 26

Check constrains

t = - 1 + 26

1 0^{2} + 8^{2} = c^{2}

- x^{2} - 4 x + 1

4 x (2 - x) = 0

(5 x + 2) (x - 1) = 7 x

t^{2} + 3 t + 2 = 0

What are the solutions to the equation t(t)=25-2t,0<= t<= 10 ?
The solutions to the equation t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)
Find the zeros of t(t)=25-2t,0<= t<= 10
The zeros of t(t)=25-2t,0<= t<= 10 are t=-1+sqrt(26)