What are the solutions to the equation 6y^2-3y-4=y^2 ?

The solutions to the equation 6y^2-3y-4=y^2 are y=(3+sqrt(89))/(10),y=(3-sqrt(89))/(10)

Find the zeros of 6y^2-3y-4=y^2

The zeros of 6y^2-3y-4=y^2 are y=(3+sqrt(89))/(10),y=(3-sqrt(89))/(10)

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6 y^{2} - 3 y - 4 = y^{2}

y = \frac{3 + 89}{10}, y = \frac{3 - 89}{10}

Decimal Notation

y = 1.24339 \dots, y = - 0.64339 \dots

Solution steps

6 y^{2} - 3 y - 4 = y^{2}

Move

y^{2}

to the left side

5 y^{2} - 3 y - 4 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 5 (- 4) = 89

y_{1, 2} = \frac{- ( - 3 ) \pm 89}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 3 ) + 89}{2 \cdot 5}, y_{2} = \frac{- ( - 3 ) - 89}{2 \cdot 5}

y = \frac{- ( - 3 ) + 89}{2 \cdot 5} : \frac{3 + 89}{10}

y = \frac{- ( - 3 ) - 89}{2 \cdot 5} : \frac{3 - 89}{10}

The solutions to the quadratic equation are:

y = \frac{3 + 89}{10}, y = \frac{3 - 89}{10}

y^{2} - 17 y + 16 = 0

6 x^{2} - 5 x - 4 = 0

(2 x - 1) (x) = 120

x^{2} + 3 x + 13 = 3 - 4 x

10 b^{2} + 21 b + 6 = 2 b

What are the solutions to the equation 6y^2-3y-4=y^2 ?
The solutions to the equation 6y^2-3y-4=y^2 are y=(3+sqrt(89))/(10),y=(3-sqrt(89))/(10)
Find the zeros of 6y^2-3y-4=y^2
The zeros of 6y^2-3y-4=y^2 are y=(3+sqrt(89))/(10),y=(3-sqrt(89))/(10)