What are the solutions to the equation 24y^2-9+12y-3=32y+3 ?

The solutions to the equation 24y^2-9+12y-3=32y+3 are y=(5+sqrt(115))/(12),y=(5-sqrt(115))/(12)

Find the zeros of 24y^2-9+12y-3=32y+3

The zeros of 24y^2-9+12y-3=32y+3 are y=(5+sqrt(115))/(12),y=(5-sqrt(115))/(12)

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24 y^{2} - 9 + 12 y - 3 = 32 y + 3

y = \frac{5 + 115}{12}, y = \frac{5 - 115}{12}

Decimal Notation

y = 1.31031 \dots, y = - 0.47698 \dots

Solution steps

24 y^{2} - 9 + 12 y - 3 = 32 y + 3

Refine

24 y^{2} + 12 y - 12 = 32 y + 3

Move

3

to the left side

24 y^{2} + 12 y - 15 = 32 y

Move

32 y

to the left side

24 y^{2} - 20 y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 24 ( - 15 )}{2 \cdot 24}

(- 20)^{2} - 4 \cdot 24 (- 15) = 4115

y_{1, 2} = \frac{- ( - 20 ) \pm 4 115}{2 \cdot 24}

Separate the solutions

y_{1} = \frac{- ( - 20 ) + 4 115}{2 \cdot 24}, y_{2} = \frac{- ( - 20 ) - 4 115}{2 \cdot 24}

y = \frac{- ( - 20 ) + 4 115}{2 \cdot 24} : \frac{5 + 115}{12}

y = \frac{- ( - 20 ) - 4 115}{2 \cdot 24} : \frac{5 - 115}{12}

The solutions to the quadratic equation are:

y = \frac{5 + 115}{12}, y = \frac{5 - 115}{12}

x, x (x - 1) = e

3 + x - 2 x^{2} = 0

\frac{3 x ^{2}}{2} = 0

5 x^{2} - 2 = 23

r^{2} = 24

What are the solutions to the equation 24y^2-9+12y-3=32y+3 ?
The solutions to the equation 24y^2-9+12y-3=32y+3 are y=(5+sqrt(115))/(12),y=(5-sqrt(115))/(12)
Find the zeros of 24y^2-9+12y-3=32y+3
The zeros of 24y^2-9+12y-3=32y+3 are y=(5+sqrt(115))/(12),y=(5-sqrt(115))/(12)