What are the solutions to the equation 25t^2-20t+4=0 ?

The solutions to the equation 25t^2-20t+4=0 are t= 2/5

Find the zeros of 25t^2-20t+4=0

The zeros of 25t^2-20t+4=0 are t= 2/5

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

t = \frac{2}{5}

Decimal Notation

t = 0.4

Solution steps

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

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 25 \cdot 4}{2 \cdot 25}

(- 20)^{2} - 4 \cdot 25 \cdot 4 = 0

t_{1, 2} = \frac{- ( - 20 ) \pm 0}{2 \cdot 25}

t = \frac{- ( - 20 )}{2 \cdot 25}

\frac{- ( - 20 )}{2 \cdot 25} = \frac{2}{5}

t = \frac{2}{5}

The solution to the quadratic equation is:

t = \frac{2}{5}

x (10 x - 13) = 3

(3 x + 2) (5 x - 1) = 4

3 t^{2} + 10 t + 1 = t

11 + 12 x^{2} = 10 - 15 x

16.1 x^{2} - 200 x - 200 = 0

What are the solutions to the equation 25t^2-20t+4=0 ?
The solutions to the equation 25t^2-20t+4=0 are t= 2/5
Find the zeros of 25t^2-20t+4=0
The zeros of 25t^2-20t+4=0 are t= 2/5