What are the solutions to the equation 4y^2+12y-3=5y ?

The solutions to the equation 4y^2+12y-3=5y are y=(-7+sqrt(97))/8 ,y=(-7-sqrt(97))/8

Find the zeros of 4y^2+12y-3=5y

The zeros of 4y^2+12y-3=5y are y=(-7+sqrt(97))/8 ,y=(-7-sqrt(97))/8

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4 y^{2} + 12 y - 3 = 5 y

y = \frac{- 7 + 97}{8}, y = \frac{- 7 - 97}{8}

Decimal Notation

y = 0.35610 \dots, y = - 2.10610 \dots

Solution steps

4 y^{2} + 12 y - 3 = 5 y

Move

5 y

to the left side

4 y^{2} + 7 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 4 ( - 3 )}{2 \cdot 4}

7^{2} - 4 \cdot 4 (- 3) = 97

y_{1, 2} = \frac{- 7 \pm 97}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 7 + 97}{2 \cdot 4}, y_{2} = \frac{- 7 - 97}{2 \cdot 4}

y = \frac{- 7 + 97}{2 \cdot 4} : \frac{- 7 + 97}{8}

y = \frac{- 7 - 97}{2 \cdot 4} : \frac{- 7 - 97}{8}

The solutions to the quadratic equation are:

y = \frac{- 7 + 97}{8}, y = \frac{- 7 - 97}{8}

2 b^{2} - 32 = 0

20 x^{2} - 7 x = 14

x^{2} - 8 x - 3 = 6

x, x^{2} + x y + y^{2} = 16

3 x^{2} + x - \frac{1}{12} = 0

What are the solutions to the equation 4y^2+12y-3=5y ?
The solutions to the equation 4y^2+12y-3=5y are y=(-7+sqrt(97))/8 ,y=(-7-sqrt(97))/8
Find the zeros of 4y^2+12y-3=5y
The zeros of 4y^2+12y-3=5y are y=(-7+sqrt(97))/8 ,y=(-7-sqrt(97))/8