What are the solutions to the equation t=-3.3t^2+5t+6 ?

The solutions to the equation t=-3.3t^2+5t+6 are t=-(2(sqrt(595)-10))/(33),t=(2(10+sqrt(595)))/(33)

Find the zeros of t=-3.3t^2+5t+6

The zeros of t=-3.3t^2+5t+6 are t=-(2(sqrt(595)-10))/(33),t=(2(10+sqrt(595)))/(33)

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t = - 3.3 t^{2} + 5 t + 6

t = - \frac{2 ( 595 - 10 )}{33}, t = \frac{2 ( 10 + 595 )}{33}

Decimal Notation

t = - 0.87228 \dots, t = 2.08440 \dots

Solution steps

t = - 3.3 t^{2} + 5 t + 6

Multiply both sides by

10

t \cdot 10 = - 33 t^{2} + 50 t + 60

Switch sides

- 33 t^{2} + 50 t + 60 = t \cdot 10

Move

t 10

to the left side

- 33 t^{2} + 40 t + 60 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 40 \pm 4 0 ^{2} - 4 ( - 33 ) \cdot 60}{2 ( - 33 )}

4 0^{2} - 4 (- 33) \cdot 60 = 4595

t_{1, 2} = \frac{- 40 \pm 4 595}{2 ( - 33 )}

Separate the solutions

t_{1} = \frac{- 40 + 4 595}{2 ( - 33 )}, t_{2} = \frac{- 40 - 4 595}{2 ( - 33 )}

t = \frac{- 40 + 4 595}{2 ( - 33 )} : - \frac{2 ( 595 - 10 )}{33}

t = \frac{- 40 - 4 595}{2 ( - 33 )} : \frac{2 ( 10 + 595 )}{33}

The solutions to the quadratic equation are:

t = - \frac{2 ( 595 - 10 )}{33}, t = \frac{2 ( 10 + 595 )}{33}

- 4 x^{2} + 28 x - 40 = 0

3 q^{2} - 136 q + 1200 = 0

40 + 3 x - 5 x^{2} = 0

0.8 x^{2} + 0.8 x - 5 = 0

c^{2} + c - 7 = 0

What are the solutions to the equation t=-3.3t^2+5t+6 ?
The solutions to the equation t=-3.3t^2+5t+6 are t=-(2(sqrt(595)-10))/(33),t=(2(10+sqrt(595)))/(33)
Find the zeros of t=-3.3t^2+5t+6
The zeros of t=-3.3t^2+5t+6 are t=-(2(sqrt(595)-10))/(33),t=(2(10+sqrt(595)))/(33)