What are the solutions to the equation 4y^2=16y+3 ?

The solutions to the equation 4y^2=16y+3 are y=(4+sqrt(19))/2 ,y=(4-sqrt(19))/2

Find the zeros of 4y^2=16y+3

The zeros of 4y^2=16y+3 are y=(4+sqrt(19))/2 ,y=(4-sqrt(19))/2

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4 y^{2} = 16 y + 3

y = \frac{4 + 19}{2}, y = \frac{4 - 19}{2}

Solution steps

4 y^{2} = 16 y + 3

Move

3

to the left side

4 y^{2} - 3 = 16 y

Move

16 y

to the left side

4 y^{2} - 3 - 16 y = 0

Write in the standard form

a x^{2} + b x + c = 0

4 y^{2} - 16 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

(- 16)^{2} - 4 \cdot 4 (- 3) = 419

y_{1, 2} = \frac{- ( - 16 ) \pm 4 19}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 16 ) + 4 19}{2 \cdot 4}, y_{2} = \frac{- ( - 16 ) - 4 19}{2 \cdot 4}

y = \frac{- ( - 16 ) + 4 19}{2 \cdot 4} : \frac{4 + 19}{2}

y = \frac{- ( - 16 ) - 4 19}{2 \cdot 4} : \frac{4 - 19}{2}

The solutions to the quadratic equation are:

y = \frac{4 + 19}{2}, y = \frac{4 - 19}{2}

6 - 2 x - x^{2} = x + 6

2 x^{2} - 4 < 0

x^{2} + 4 x = - 5

9 e = 45

\frac{27 x ^{6} y ^{3}}{9 x ^{4} y ^{5}}

What are the solutions to the equation 4y^2=16y+3 ?
The solutions to the equation 4y^2=16y+3 are y=(4+sqrt(19))/2 ,y=(4-sqrt(19))/2
Find the zeros of 4y^2=16y+3
The zeros of 4y^2=16y+3 are y=(4+sqrt(19))/2 ,y=(4-sqrt(19))/2