What are the solutions to the equation 16x^2-80x+64=0 ?

The solutions to the equation 16x^2-80x+64=0 are x=4,x=1

Find the zeros of 16x^2-80x+64=0

The zeros of 16x^2-80x+64=0 are x=4,x=1

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16 x^{2} - 80 x + 64 = 0

x = 4, x = 1

Solution steps

16 x^{2} - 80 x + 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 \cdot 16 \cdot 64}{2 \cdot 16}

(- 80)^{2} - 4 \cdot 16 \cdot 64 = 48

x_{1, 2} = \frac{- ( - 80 ) \pm 48}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 80 ) + 48}{2 \cdot 16}, x_{2} = \frac{- ( - 80 ) - 48}{2 \cdot 16}

x = \frac{- ( - 80 ) + 48}{2 \cdot 16} : 4

x = \frac{- ( - 80 ) - 48}{2 \cdot 16} : 1

The solutions to the quadratic equation are:

x = 4, x = 1

(x - 5)^{2} = (x + 5)^{2} - 3 x^{2} - 25

4 x^{2} = 28 x - 49

3 m^{2} - 9 m - 11 = 0

(x) (\frac{0.5 x}{0.939}) = 80

r^{2} = 2 (1 - r) r

What are the solutions to the equation 16x^2-80x+64=0 ?
The solutions to the equation 16x^2-80x+64=0 are x=4,x=1
Find the zeros of 16x^2-80x+64=0
The zeros of 16x^2-80x+64=0 are x=4,x=1