What are the solutions to the equation t^2-5t-5=-2t ?

The solutions to the equation t^2-5t-5=-2t are t=(3+sqrt(29))/2 ,t=(3-sqrt(29))/2

Find the zeros of t^2-5t-5=-2t

The zeros of t^2-5t-5=-2t are t=(3+sqrt(29))/2 ,t=(3-sqrt(29))/2

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t^{2} - 5 t - 5 = - 2 t

t = \frac{3 + 29}{2}, t = \frac{3 - 29}{2}

Decimal Notation

t = 4.19258 \dots, t = - 1.19258 \dots

Solution steps

t^{2} - 5 t - 5 = - 2 t

Move

2 t

to the left side

t^{2} - 3 t - 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 5 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 5) = 29

t_{1, 2} = \frac{- ( - 3 ) \pm 29}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 3 ) + 29}{2 \cdot 1}, t_{2} = \frac{- ( - 3 ) - 29}{2 \cdot 1}

t = \frac{- ( - 3 ) + 29}{2 \cdot 1} : \frac{3 + 29}{2}

t = \frac{- ( - 3 ) - 29}{2 \cdot 1} : \frac{3 - 29}{2}

The solutions to the quadratic equation are:

t = \frac{3 + 29}{2}, t = \frac{3 - 29}{2}

- 2 x^{2} - 2 x - 4 = 0

2 x^{2} + x - 1

8 x^{2} - 33 x + 4 = 0

x - x^{2} = \frac{1}{6}

6 x^{2} = 7 x + 2

What are the solutions to the equation t^2-5t-5=-2t ?
The solutions to the equation t^2-5t-5=-2t are t=(3+sqrt(29))/2 ,t=(3-sqrt(29))/2
Find the zeros of t^2-5t-5=-2t
The zeros of t^2-5t-5=-2t are t=(3+sqrt(29))/2 ,t=(3-sqrt(29))/2