What are the solutions to the equation t^2-12t+36=0 ?

The solutions to the equation t^2-12t+36=0 are t=6

Find the zeros of t^2-12t+36=0

The zeros of t^2-12t+36=0 are t=6

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t^{2} - 12 t + 36 = 0

t = 6

Solution steps

t^{2} - 12 t + 36 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot 36 = 0

t_{1, 2} = \frac{- ( - 12 ) \pm 0}{2 \cdot 1}

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

\frac{- ( - 12 )}{2 \cdot 1} = 6

t = 6

The solution to the quadratic equation is:

t = 6

7 x^{2} - x - 12 = 0

(x + 2) (x + 4) = 3

x^{2} - 11 x + 30 = 0

4 r^{2} - 20 r + 25 = 0

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

What are the solutions to the equation t^2-12t+36=0 ?
The solutions to the equation t^2-12t+36=0 are t=6
Find the zeros of t^2-12t+36=0
The zeros of t^2-12t+36=0 are t=6