What are the solutions to the equation 2y^2+8y-6=3y ?

The solutions to the equation 2y^2+8y-6=3y are y=(-5+sqrt(73))/4 ,y=(-5-sqrt(73))/4

Find the zeros of 2y^2+8y-6=3y

The zeros of 2y^2+8y-6=3y are y=(-5+sqrt(73))/4 ,y=(-5-sqrt(73))/4

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2 y^{2} + 8 y - 6 = 3 y

y = \frac{- 5 + 73}{4}, y = \frac{- 5 - 73}{4}

Decimal Notation

y = 0.88600 \dots, y = - 3.38600 \dots

Solution steps

2 y^{2} + 8 y - 6 = 3 y

Move

3 y

to the left side

2 y^{2} + 5 y - 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 2 ( - 6 )}{2 \cdot 2}

5^{2} - 4 \cdot 2 (- 6) = 73

y_{1, 2} = \frac{- 5 \pm 73}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 5 + 73}{2 \cdot 2}, y_{2} = \frac{- 5 - 73}{2 \cdot 2}

y = \frac{- 5 + 73}{2 \cdot 2} : \frac{- 5 + 73}{4}

y = \frac{- 5 - 73}{2 \cdot 2} : \frac{- 5 - 73}{4}

The solutions to the quadratic equation are:

y = \frac{- 5 + 73}{4}, y = \frac{- 5 - 73}{4}

- 4 x - 7 \geq 21

5.5 x + 6.25 (100) = 6 (x + 100)

x^{2} + 10 x \geq - 24

\frac{π}{3} - \frac{π}{3}

- 2 (2 x - 4) = 2 x

What are the solutions to the equation 2y^2+8y-6=3y ?
The solutions to the equation 2y^2+8y-6=3y are y=(-5+sqrt(73))/4 ,y=(-5-sqrt(73))/4
Find the zeros of 2y^2+8y-6=3y
The zeros of 2y^2+8y-6=3y are y=(-5+sqrt(73))/4 ,y=(-5-sqrt(73))/4