What are the solutions to the equation 5y^2+7y-3=4y ?

The solutions to the equation 5y^2+7y-3=4y are y=(-3+sqrt(69))/(10),y=(-3-sqrt(69))/(10)

Find the zeros of 5y^2+7y-3=4y

The zeros of 5y^2+7y-3=4y are y=(-3+sqrt(69))/(10),y=(-3-sqrt(69))/(10)

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

y = \frac{- 3 + 69}{10}, y = \frac{- 3 - 69}{10}

Decimal Notation

y = 0.53066 \dots, y = - 1.13066 \dots

Solution steps

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

Move

4 y

to the left side

5 y^{2} + 3 y - 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 5 ( - 3 )}{2 \cdot 5}

3^{2} - 4 \cdot 5 (- 3) = 69

y_{1, 2} = \frac{- 3 \pm 69}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- 3 + 69}{2 \cdot 5}, y_{2} = \frac{- 3 - 69}{2 \cdot 5}

y = \frac{- 3 + 69}{2 \cdot 5} : \frac{- 3 + 69}{10}

y = \frac{- 3 - 69}{2 \cdot 5} : \frac{- 3 - 69}{10}

The solutions to the quadratic equation are:

y = \frac{- 3 + 69}{10}, y = \frac{- 3 - 69}{10}

x^{2} = \frac{3}{250}

x^{2} - 10 x + 36 = 0

u^{2} + 3 u + 2 = 0

x^{2} + 9 x = 0

72 = 2 π + r^{2}

What are the solutions to the equation 5y^2+7y-3=4y ?
The solutions to the equation 5y^2+7y-3=4y are y=(-3+sqrt(69))/(10),y=(-3-sqrt(69))/(10)
Find the zeros of 5y^2+7y-3=4y
The zeros of 5y^2+7y-3=4y are y=(-3+sqrt(69))/(10),y=(-3-sqrt(69))/(10)