What are the solutions to the equation 16x^2-32x+16=1 ?

The solutions to the equation 16x^2-32x+16=1 are x= 5/4 ,x= 3/4

Find the zeros of 16x^2-32x+16=1

The zeros of 16x^2-32x+16=1 are x= 5/4 ,x= 3/4

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16 x^{2} - 32 x + 16 = 1

x = \frac{5}{4}, x = \frac{3}{4}

Solution steps

16 x^{2} - 32 x + 16 = 1

Move

1

to the left side

16 x^{2} - 32 x + 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 16 \cdot 15}{2 \cdot 16}

(- 32)^{2} - 4 \cdot 16 \cdot 15 = 8

x_{1, 2} = \frac{- ( - 32 ) \pm 8}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 32 ) + 8}{2 \cdot 16}, x_{2} = \frac{- ( - 32 ) - 8}{2 \cdot 16}

x = \frac{- ( - 32 ) + 8}{2 \cdot 16} : \frac{5}{4}

x = \frac{- ( - 32 ) - 8}{2 \cdot 16} : \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{5}{4}, x = \frac{3}{4}

a^{2} + 2 a - 1 = 0

x^{2} + 2 x - 7 = 17

- 3 a^{2} - 8 a - 5 = 0

x^{2} + 0 x - 4 = 0

n^{2} - 4 n = 11

What are the solutions to the equation 16x^2-32x+16=1 ?
The solutions to the equation 16x^2-32x+16=1 are x= 5/4 ,x= 3/4
Find the zeros of 16x^2-32x+16=1
The zeros of 16x^2-32x+16=1 are x= 5/4 ,x= 3/4