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

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

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

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

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

t = 5, t = 1

Solution steps

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

Solve with the quadratic formula

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

(- 18)^{2} - 4 \cdot 3 \cdot 15 = 12

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 5, t = 1

1 0^{2} + 1 6^{2} = x^{2}

a^{n} + a^{n + 1} + a^{n + 2}

6 n - 12 = - 18

a^{\frac{1}{2}} a^{\frac{1}{3}}

5 x^{2} - 3 x - 7 = 0

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