What are the solutions to the equation -3t^2+12t-9=0 ?

The solutions to the equation -3t^2+12t-9=0 are t=1,t=3

Find the zeros of -3t^2+12t-9=0

The zeros of -3t^2+12t-9=0 are t=1,t=3

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- 3 t^{2} + 12 t - 9 = 0

t = 1, t = 3

Solution steps

- 3 t^{2} + 12 t - 9 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) ( - 9 )}{2 ( - 3 )}

1 2^{2} - 4 (- 3) (- 9) = 6

t_{1, 2} = \frac{- 12 \pm 6}{2 ( - 3 )}

Separate the solutions

t_{1} = \frac{- 12 + 6}{2 ( - 3 )}, t_{2} = \frac{- 12 - 6}{2 ( - 3 )}

t = \frac{- 12 + 6}{2 ( - 3 )} : 1

t = \frac{- 12 - 6}{2 ( - 3 )} : 3

The solutions to the quadratic equation are:

t = 1, t = 3

2 r^{2} + 3 r + 1 = 0

\frac{1}{25} - x^{2} = 0

3 x^{2} + 6 = 11 x

x^{2} + (x - 3)^{2} = 549

m^{2} - 8 m - 20 = 0

What are the solutions to the equation -3t^2+12t-9=0 ?
The solutions to the equation -3t^2+12t-9=0 are t=1,t=3
Find the zeros of -3t^2+12t-9=0
The zeros of -3t^2+12t-9=0 are t=1,t=3