What are the solutions to the equation -2t^2+9t+3=-6t^2 ?

The solutions to the equation -2t^2+9t+3=-6t^2 are t=(-9+sqrt(33))/8 ,t=(-9-sqrt(33))/8

Find the zeros of -2t^2+9t+3=-6t^2

The zeros of -2t^2+9t+3=-6t^2 are t=(-9+sqrt(33))/8 ,t=(-9-sqrt(33))/8

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- 2 t^{2} + 9 t + 3 = - 6 t^{2}

t = \frac{- 9 + 33}{8}, t = \frac{- 9 - 33}{8}

Decimal Notation

t = - 0.40692 \dots, t = - 1.84307 \dots

Solution steps

- 2 t^{2} + 9 t + 3 = - 6 t^{2}

Move

6 t^{2}

to the left side

4 t^{2} + 9 t + 3 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 9 \pm 9 ^{2} - 4 \cdot 4 \cdot 3}{2 \cdot 4}

9^{2} - 4 \cdot 4 \cdot 3 = 33

t_{1, 2} = \frac{- 9 \pm 33}{2 \cdot 4}

Separate the solutions

t_{1} = \frac{- 9 + 33}{2 \cdot 4}, t_{2} = \frac{- 9 - 33}{2 \cdot 4}

t = \frac{- 9 + 33}{2 \cdot 4} : \frac{- 9 + 33}{8}

t = \frac{- 9 - 33}{2 \cdot 4} : \frac{- 9 - 33}{8}

The solutions to the quadratic equation are:

t = \frac{- 9 + 33}{8}, t = \frac{- 9 - 33}{8}

(x - 10)^{2} = 25

25 x^{2} = 9

4 x^{2} - 12 x + 9 = 9

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

2 x^{2} - 2 x - 15 = 0

What are the solutions to the equation -2t^2+9t+3=-6t^2 ?
The solutions to the equation -2t^2+9t+3=-6t^2 are t=(-9+sqrt(33))/8 ,t=(-9-sqrt(33))/8
Find the zeros of -2t^2+9t+3=-6t^2
The zeros of -2t^2+9t+3=-6t^2 are t=(-9+sqrt(33))/8 ,t=(-9-sqrt(33))/8