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

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

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

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

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

t = \frac{5}{2}

Decimal Notation

t = 2.5

Solution steps

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

Solve with the quadratic formula

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

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

t_{1, 2} = \frac{- ( - 20 ) \pm 0}{2 \cdot 4}

t = \frac{- ( - 20 )}{2 \cdot 4}

\frac{- ( - 20 )}{2 \cdot 4} = \frac{5}{2}

t = \frac{5}{2}

The solution to the quadratic equation is:

t = \frac{5}{2}

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

9 x^{2} + 1 = 5

(x - 5) (3 x + 4) = 0

4 x^{2} + 12 x = - 17

(6 - R)^{2} + (3 - R)^{2} = R^{2}

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