What are the solutions to the equation 19y^2-13y+2=-y^2 ?

The solutions to the equation 19y^2-13y+2=-y^2 are y= 2/5 ,y= 1/4

Find the zeros of 19y^2-13y+2=-y^2

The zeros of 19y^2-13y+2=-y^2 are y= 2/5 ,y= 1/4

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19 y^{2} - 13 y + 2 = - y^{2}

y = \frac{2}{5}, y = \frac{1}{4}

Decimal Notation

y = 0.4, y = 0.25

Solution steps

19 y^{2} - 13 y + 2 = - y^{2}

Move

y^{2}

to the left side

20 y^{2} - 13 y + 2 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 13 ) \pm ( - 13 ) ^{2} - 4 \cdot 20 \cdot 2}{2 \cdot 20}

(- 13)^{2} - 4 \cdot 20 \cdot 2 = 3

y_{1, 2} = \frac{- ( - 13 ) \pm 3}{2 \cdot 20}

Separate the solutions

y_{1} = \frac{- ( - 13 ) + 3}{2 \cdot 20}, y_{2} = \frac{- ( - 13 ) - 3}{2 \cdot 20}

y = \frac{- ( - 13 ) + 3}{2 \cdot 20} : \frac{2}{5}

y = \frac{- ( - 13 ) - 3}{2 \cdot 20} : \frac{1}{4}

The solutions to the quadratic equation are:

y = \frac{2}{5}, y = \frac{1}{4}

- 5 (3 m + 6) = - 3 (4 m - 2)

25 x^{2} + 20 x + 4

\frac{1}{3} (1) - 1

\frac{5}{4} (x - 10) \leq 10

(- 2)^{2} + 3

What are the solutions to the equation 19y^2-13y+2=-y^2 ?
The solutions to the equation 19y^2-13y+2=-y^2 are y= 2/5 ,y= 1/4
Find the zeros of 19y^2-13y+2=-y^2
The zeros of 19y^2-13y+2=-y^2 are y= 2/5 ,y= 1/4