What are the solutions to the equation t(t)=25-2t ?

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

Find the zeros of t(t)=25-2t

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

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t (t) = 25 - 2 t

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

Decimal Notation

t = 4.09901 \dots, t = - 6.09901 \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

2 + 3 (2 x + 1) (2 x - 1) = 3

10 - 4 x - x^{2}

x^{2} - 4 x - 20 = - 5 x

(2 - x)^{2} - (2 x - 3)^{2} = 50

(5 - x)^{2} - (3 x - 1)^{2} = 0

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