What are the solutions to the equation t^2-15=-2t ?

The solutions to the equation t^2-15=-2t are t=3,t=-5

Find the zeros of t^2-15=-2t

The zeros of t^2-15=-2t are t=3,t=-5

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t^{2} - 15 = - 2 t

t = 3, t = - 5

Solution steps

t^{2} - 15 = - 2 t

Move

2 t

to the left side

t^{2} - 15 + 2 t = 0

Write in the standard form

a x^{2} + b x + c = 0

t^{2} + 2 t - 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 15) = 8

t_{1, 2} = \frac{- 2 \pm 8}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- 2 + 8}{2 \cdot 1}, t_{2} = \frac{- 2 - 8}{2 \cdot 1}

t = \frac{- 2 + 8}{2 \cdot 1} : 3

t = \frac{- 2 - 8}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

t = 3, t = - 5

3 x^{2} + 18 x = 21

x^{2} - 6 x + 19 = 0

2 (x + 5) = - 8

\frac{8 !}{3 ! 5 !}

2 x^{2} + 15 x + 27

What are the solutions to the equation t^2-15=-2t ?
The solutions to the equation t^2-15=-2t are t=3,t=-5
Find the zeros of t^2-15=-2t
The zeros of t^2-15=-2t are t=3,t=-5