What are the solutions to the equation t^2-10t+25=0 ?

The solutions to the equation t^2-10t+25=0 are t=5

Find the zeros of t^2-10t+25=0

The zeros of t^2-10t+25=0 are t=5

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t^{2} - 10 t + 25 = 0

t = 5

Solution steps

t^{2} - 10 t + 25 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 1 \cdot 25}{2 \cdot 1}

(- 10)^{2} - 4 \cdot 1 \cdot 25 = 0

t_{1, 2} = \frac{- ( - 10 ) \pm 0}{2 \cdot 1}

t = \frac{- ( - 10 )}{2 \cdot 1}

\frac{- ( - 10 )}{2 \cdot 1} = 5

t = 5

The solution to the quadratic equation is:

t = 5

lo g_{10} (3)

4 a^{2} - 12 a + 9

\frac{8}{3}

s^{2} + 6 s - 16 = 0

\frac{3}{3}

What are the solutions to the equation t^2-10t+25=0 ?
The solutions to the equation t^2-10t+25=0 are t=5
Find the zeros of t^2-10t+25=0
The zeros of t^2-10t+25=0 are t=5