What are the solutions to the equation 15+4y^2=17y ?

The solutions to the equation 15+4y^2=17y are y=3,y= 5/4

Find the zeros of 15+4y^2=17y

The zeros of 15+4y^2=17y are y=3,y= 5/4

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15 + 4 y^{2} = 17 y

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

Decimal Notation

y = 3, y = 1.25

Solution steps

15 + 4 y^{2} = 17 y

Move

17 y

to the left side

15 + 4 y^{2} - 17 y = 0

Write in the standard form

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

4 y^{2} - 17 y + 15 = 0

Solve with the quadratic formula

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

(- 17)^{2} - 4 \cdot 4 \cdot 15 = 7

y_{1, 2} = \frac{- ( - 17 ) \pm 7}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- ( - 17 ) + 7}{2 \cdot 4}, y_{2} = \frac{- ( - 17 ) - 7}{2 \cdot 4}

y = \frac{- ( - 17 ) + 7}{2 \cdot 4} : 3

y = \frac{- ( - 17 ) - 7}{2 \cdot 4} : \frac{5}{4}

The solutions to the quadratic equation are:

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

12 x^{2} + 18 x - 12 = 0

(5 b + 1) (6 b + 1) = 0

t, y = - 14 t^{2} + V o t

36 x^{2} - 396 x + 1080 = 0

1.06 = (1 + x)^{2}

What are the solutions to the equation 15+4y^2=17y ?
The solutions to the equation 15+4y^2=17y are y=3,y= 5/4
Find the zeros of 15+4y^2=17y
The zeros of 15+4y^2=17y are y=3,y= 5/4