What are the solutions to the equation 2y^2-19y+17=-3y^2 ?

The solutions to the equation 2y^2-19y+17=-3y^2 are y=(19+sqrt(21))/(10),y=(19-sqrt(21))/(10)

Find the zeros of 2y^2-19y+17=-3y^2

The zeros of 2y^2-19y+17=-3y^2 are y=(19+sqrt(21))/(10),y=(19-sqrt(21))/(10)

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2 y^{2} - 19 y + 17 = - 3 y^{2}

y = \frac{19 + 21}{10}, y = \frac{19 - 21}{10}

Decimal Notation

y = 2.35825 \dots, y = 1.44174 \dots

Solution steps

2 y^{2} - 19 y + 17 = - 3 y^{2}

Move

3 y^{2}

to the left side

5 y^{2} - 19 y + 17 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 19 ) \pm ( - 19 ) ^{2} - 4 \cdot 5 \cdot 17}{2 \cdot 5}

(- 19)^{2} - 4 \cdot 5 \cdot 17 = 21

y_{1, 2} = \frac{- ( - 19 ) \pm 21}{2 \cdot 5}

Separate the solutions

y_{1} = \frac{- ( - 19 ) + 21}{2 \cdot 5}, y_{2} = \frac{- ( - 19 ) - 21}{2 \cdot 5}

y = \frac{- ( - 19 ) + 21}{2 \cdot 5} : \frac{19 + 21}{10}

y = \frac{- ( - 19 ) - 21}{2 \cdot 5} : \frac{19 - 21}{10}

The solutions to the quadratic equation are:

y = \frac{19 + 21}{10}, y = \frac{19 - 21}{10}

x^{2} + 20 x - 3500 = 0

0 = 2 (x + 1)^{2} + 3

y, x^{2} + 2 x y - y^{2} + x = 2

x^{2} + 7 x = 0

4 x^{2} - 11 = 12 x

What are the solutions to the equation 2y^2-19y+17=-3y^2 ?
The solutions to the equation 2y^2-19y+17=-3y^2 are y=(19+sqrt(21))/(10),y=(19-sqrt(21))/(10)
Find the zeros of 2y^2-19y+17=-3y^2
The zeros of 2y^2-19y+17=-3y^2 are y=(19+sqrt(21))/(10),y=(19-sqrt(21))/(10)