What are the solutions to the equation -3t^2+18t-15=0 ?

The solutions to the equation -3t^2+18t-15=0 are t=1,t=5

Find the zeros of -3t^2+18t-15=0

The zeros of -3t^2+18t-15=0 are t=1,t=5

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- 3 t^{2} + 18 t - 15 = 0

t = 1, t = 5

Solution steps

- 3 t^{2} + 18 t - 15 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 3 ) ( - 15 )}{2 ( - 3 )}

1 8^{2} - 4 (- 3) (- 15) = 12

t_{1, 2} = \frac{- 18 \pm 12}{2 ( - 3 )}

Separate the solutions

t_{1} = \frac{- 18 + 12}{2 ( - 3 )}, t_{2} = \frac{- 18 - 12}{2 ( - 3 )}

t = \frac{- 18 + 12}{2 ( - 3 )} : 1

t = \frac{- 18 - 12}{2 ( - 3 )} : 5

The solutions to the quadratic equation are:

t = 1, t = 5

x, 3 x^{2} + 2 y^{2} = z

3 x^{2} - 6 x + 3 = 0

x^{2} + 5^{2} = 1 2^{2}

36 x^{2} - 24 x + 4 = 9

x + 23 = x^{2} + 6 x + 9

What are the solutions to the equation -3t^2+18t-15=0 ?
The solutions to the equation -3t^2+18t-15=0 are t=1,t=5
Find the zeros of -3t^2+18t-15=0
The zeros of -3t^2+18t-15=0 are t=1,t=5