What are the solutions to the equation -5(t^2-3)+15=-2t ?

The solutions to the equation -5(t^2-3)+15=-2t are t=-(-1+sqrt(151))/5 ,t=(1+sqrt(151))/5

Find the zeros of -5(t^2-3)+15=-2t

The zeros of -5(t^2-3)+15=-2t are t=-(-1+sqrt(151))/5 ,t=(1+sqrt(151))/5

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- 5 (t^{2} - 3) + 15 = - 2 t

t = - \frac{- 1 + 151}{5}, t = \frac{1 + 151}{5}

Decimal Notation

t = - 2.25764 \dots, t = 2.65764 \dots

Solution steps

- 5 (t^{2} - 3) + 15 = - 2 t

Expand

- 5 (t^{2} - 3) + 15 : - 5 t^{2} + 30

- 5 t^{2} + 30 = - 2 t

Move

2 t

to the left side

- 5 t^{2} + 30 + 2 t = 0

Write in the standard form

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

- 5 t^{2} + 2 t + 30 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 5 ) \cdot 30}{2 ( - 5 )}

2^{2} - 4 (- 5) \cdot 30 = 2151

t_{1, 2} = \frac{- 2 \pm 2 151}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 2 + 2 151}{2 ( - 5 )}, t_{2} = \frac{- 2 - 2 151}{2 ( - 5 )}

t = \frac{- 2 + 2 151}{2 ( - 5 )} : - \frac{- 1 + 151}{5}

t = \frac{- 2 - 2 151}{2 ( - 5 )} : \frac{1 + 151}{5}

The solutions to the quadratic equation are:

t = - \frac{- 1 + 151}{5}, t = \frac{1 + 151}{5}

x^{2} + 6 x - 21 = 0

x^{2} = 15 - 2 x

- x + 4 x (2 - x) = - 1

l (x) = - x^{2} + 12 x - 20

x^{2} + 1.4 = 3.6

What are the solutions to the equation -5(t^2-3)+15=-2t ?
The solutions to the equation -5(t^2-3)+15=-2t are t=-(-1+sqrt(151))/5 ,t=(1+sqrt(151))/5
Find the zeros of -5(t^2-3)+15=-2t
The zeros of -5(t^2-3)+15=-2t are t=-(-1+sqrt(151))/5 ,t=(1+sqrt(151))/5