What are the solutions to the equation 4n^2-20n+16=0 ?

The solutions to the equation 4n^2-20n+16=0 are n=4,n=1

Find the zeros of 4n^2-20n+16=0

The zeros of 4n^2-20n+16=0 are n=4,n=1

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4 n^{2} - 20 n + 16 = 0

n = 4, n = 1

Solution steps

4 n^{2} - 20 n + 16 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 16 = 12

n_{1, 2} = \frac{- ( - 20 ) \pm 12}{2 \cdot 4}

Separate the solutions

n_{1} = \frac{- ( - 20 ) + 12}{2 \cdot 4}, n_{2} = \frac{- ( - 20 ) - 12}{2 \cdot 4}

n = \frac{- ( - 20 ) + 12}{2 \cdot 4} : 4

n = \frac{- ( - 20 ) - 12}{2 \cdot 4} : 1

The solutions to the quadratic equation are:

n = 4, n = 1

220 - (16.5 - 2 x)^{2} = 99.9

5 x^{2} + 20 x

8 z^{2} - 3 z - 2 = 0

x (2 x - 9) = - 10

m^{2} + 12 m = 52

What are the solutions to the equation 4n^2-20n+16=0 ?
The solutions to the equation 4n^2-20n+16=0 are n=4,n=1
Find the zeros of 4n^2-20n+16=0
The zeros of 4n^2-20n+16=0 are n=4,n=1