What are the solutions to the equation 16y^2-32y=48 ?

The solutions to the equation 16y^2-32y=48 are y=3,y=-1

Find the zeros of 16y^2-32y=48

The zeros of 16y^2-32y=48 are y=3,y=-1

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16 y^{2} - 32 y = 48

y = 3, y = - 1

Solution steps

16 y^{2} - 32 y = 48

Move

48

to the left side

16 y^{2} - 32 y - 48 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 \cdot 16 ( - 48 )}{2 \cdot 16}

(- 32)^{2} - 4 \cdot 16 (- 48) = 64

y_{1, 2} = \frac{- ( - 32 ) \pm 64}{2 \cdot 16}

Separate the solutions

y_{1} = \frac{- ( - 32 ) + 64}{2 \cdot 16}, y_{2} = \frac{- ( - 32 ) - 64}{2 \cdot 16}

y = \frac{- ( - 32 ) + 64}{2 \cdot 16} : 3

y = \frac{- ( - 32 ) - 64}{2 \cdot 16} : - 1

The solutions to the quadratic equation are:

y = 3, y = - 1

\frac{4}{3 i}

x^{4} - x^{2}

(4 - 2 i) (4 + 2 i)

5 x^{2} + x - 4

3 x - 1 \geq 11

What are the solutions to the equation 16y^2-32y=48 ?
The solutions to the equation 16y^2-32y=48 are y=3,y=-1
Find the zeros of 16y^2-32y=48
The zeros of 16y^2-32y=48 are y=3,y=-1