What are the solutions to the equation 16(y-3)^2+1=8(y-3) ?

The solutions to the equation 16(y-3)^2+1=8(y-3) are y= 13/4

Find the zeros of 16(y-3)^2+1=8(y-3)

The zeros of 16(y-3)^2+1=8(y-3) are y= 13/4

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16 (y - 3)^{2} + 1 = 8 (y - 3)

y = \frac{13}{4}

Decimal Notation

y = 3.25

Solution steps

16 (y - 3)^{2} + 1 = 8 (y - 3)

Expand

16 (y - 3)^{2} + 1 : 16 y^{2} - 96 y + 145

Expand

8 (y - 3) : 8 y - 24

16 y^{2} - 96 y + 145 = 8 y - 24

Move

24

to the left side

16 y^{2} - 96 y + 169 = 8 y

Move

8 y

to the left side

16 y^{2} - 104 y + 169 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 104 ) \pm ( - 104 ) ^{2} - 4 \cdot 16 \cdot 169}{2 \cdot 16}

(- 104)^{2} - 4 \cdot 16 \cdot 169 = 0

y_{1, 2} = \frac{- ( - 104 ) \pm 0}{2 \cdot 16}

y = \frac{- ( - 104 )}{2 \cdot 16}

\frac{- ( - 104 )}{2 \cdot 16} = \frac{13}{4}

y = \frac{13}{4}

The solution to the quadratic equation is:

y = \frac{13}{4}

2401

- 3 x \leq 15

4 \cdot 1 0^{- 7} = - 7 \cdot 1 0^{- 9} (x - 30)

∣ 3 x + 1 ∣ \geq 2 ∣ x - 6 ∣

∣ 2 x + 8 ∣ - 6 = - 12

What are the solutions to the equation 16(y-3)^2+1=8(y-3) ?
The solutions to the equation 16(y-3)^2+1=8(y-3) are y= 13/4
Find the zeros of 16(y-3)^2+1=8(y-3)
The zeros of 16(y-3)^2+1=8(y-3) are y= 13/4