What are the solutions to the equation 11y^2-19y-10=-4y^2 ?

The solutions to the equation 11y^2-19y-10=-4y^2 are y= 5/3 ,y=-2/5

Find the zeros of 11y^2-19y-10=-4y^2

The zeros of 11y^2-19y-10=-4y^2 are y= 5/3 ,y=-2/5

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11 y^{2} - 19 y - 10 = - 4 y^{2}

y = \frac{5}{3}, y = - \frac{2}{5}

Decimal Notation

y = 1.66666 \dots, y = - 0.4

Solution steps

11 y^{2} - 19 y - 10 = - 4 y^{2}

Move

4 y^{2}

to the left side

15 y^{2} - 19 y - 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 15 ( - 10 )}{2 \cdot 15}

(- 19)^{2} - 4 \cdot 15 (- 10) = 31

y_{1, 2} = \frac{- ( - 19 ) \pm 31}{2 \cdot 15}

Separate the solutions

y_{1} = \frac{- ( - 19 ) + 31}{2 \cdot 15}, y_{2} = \frac{- ( - 19 ) - 31}{2 \cdot 15}

y = \frac{- ( - 19 ) + 31}{2 \cdot 15} : \frac{5}{3}

y = \frac{- ( - 19 ) - 31}{2 \cdot 15} : - \frac{2}{5}

The solutions to the quadratic equation are:

y = \frac{5}{3}, y = - \frac{2}{5}

- 1 2^{2} + x^{2} = 1 3^{2}

6 x^{2} + 15 x - 36

\frac{2 x + 1}{3} + x = 2

4 (a + 3) = 2 a + 12

4 y - 8 (- 3) = 20

What are the solutions to the equation 11y^2-19y-10=-4y^2 ?
The solutions to the equation 11y^2-19y-10=-4y^2 are y= 5/3 ,y=-2/5
Find the zeros of 11y^2-19y-10=-4y^2
The zeros of 11y^2-19y-10=-4y^2 are y= 5/3 ,y=-2/5