What are the solutions to the equation 16r^2-40r+25=0 ?

The solutions to the equation 16r^2-40r+25=0 are r= 5/4

Find the zeros of 16r^2-40r+25=0

The zeros of 16r^2-40r+25=0 are r= 5/4

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16 r^{2} - 40 r + 25 = 0

r = \frac{5}{4}

Decimal Notation

r = 1.25

Solution steps

16 r^{2} - 40 r + 25 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 \cdot 16 \cdot 25}{2 \cdot 16}

(- 40)^{2} - 4 \cdot 16 \cdot 25 = 0

r_{1, 2} = \frac{- ( - 40 ) \pm 0}{2 \cdot 16}

r = \frac{- ( - 40 )}{2 \cdot 16}

\frac{- ( - 40 )}{2 \cdot 16} = \frac{5}{4}

r = \frac{5}{4}

The solution to the quadratic equation is:

r = \frac{5}{4}

x^{2} - 3 x - 7 < 3

3 \cdot 49

3 x + 1 - 2 x = 9 - 3

5^{0} \cdot 5^{3}

- \frac{5}{3} \cdot 3

What are the solutions to the equation 16r^2-40r+25=0 ?
The solutions to the equation 16r^2-40r+25=0 are r= 5/4
Find the zeros of 16r^2-40r+25=0
The zeros of 16r^2-40r+25=0 are r= 5/4