What are the solutions to the equation t^2-15t+54=0 ?

The solutions to the equation t^2-15t+54=0 are t=9,t=6

Find the zeros of t^2-15t+54=0

The zeros of t^2-15t+54=0 are t=9,t=6

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t^{2} - 15 t + 54 = 0

t = 9, t = 6

Solution steps

t^{2} - 15 t + 54 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 15 ) \pm ( - 15 ) ^{2} - 4 \cdot 1 \cdot 54}{2 \cdot 1}

(- 15)^{2} - 4 \cdot 1 \cdot 54 = 3

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

Separate the solutions

t_{1} = \frac{- ( - 15 ) + 3}{2 \cdot 1}, t_{2} = \frac{- ( - 15 ) - 3}{2 \cdot 1}

t = \frac{- ( - 15 ) + 3}{2 \cdot 1} : 9

t = \frac{- ( - 15 ) - 3}{2 \cdot 1} : 6

The solutions to the quadratic equation are:

t = 9, t = 6

4 x^{2} + 8 x + 0 = 0

x^{2} = 1 - \frac{9}{25}

6 x^{2} - 7 x - 20 = 0

(x - 18)^{2} = 17

4 x^{2} + 4 = 17 x

What are the solutions to the equation t^2-15t+54=0 ?
The solutions to the equation t^2-15t+54=0 are t=9,t=6
Find the zeros of t^2-15t+54=0
The zeros of t^2-15t+54=0 are t=9,t=6