What are the solutions to the equation 4y^2=15y+4 ?

The solutions to the equation 4y^2=15y+4 are y=4,y=-1/4

Find the zeros of 4y^2=15y+4

The zeros of 4y^2=15y+4 are y=4,y=-1/4

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4 y^{2} = 15 y + 4

y = 4, y = - \frac{1}{4}

Decimal Notation

y = 4, y = - 0.25

Solution steps

4 y^{2} = 15 y + 4

Move

4

to the left side

4 y^{2} - 4 = 15 y

Move

15 y

to the left side

4 y^{2} - 4 - 15 y = 0

Write in the standard form

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

4 y^{2} - 15 y - 4 = 0

Solve with the quadratic formula

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

(- 15)^{2} - 4 \cdot 4 (- 4) = 17

y_{1, 2} = \frac{- ( - 15 ) \pm 17}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 15 ) + 17}{2 \cdot 4}, y_{2} = \frac{- ( - 15 ) - 17}{2 \cdot 4}

y = \frac{- ( - 15 ) + 17}{2 \cdot 4} : 4

y = \frac{- ( - 15 ) - 17}{2 \cdot 4} : - \frac{1}{4}

The solutions to the quadratic equation are:

y = 4, y = - \frac{1}{4}

1010 \cdot 412

- \frac{3}{10} + \frac{3}{5}

\frac{2 x}{3} - \frac{x - 1}{4} < \frac{x - 1}{8}

(x^{- 3})^{- 5} x^{6}

4 x^{2} + 3 x - 3 = 0

What are the solutions to the equation 4y^2=15y+4 ?
The solutions to the equation 4y^2=15y+4 are y=4,y=-1/4
Find the zeros of 4y^2=15y+4
The zeros of 4y^2=15y+4 are y=4,y=-1/4