What are the solutions to the equation -y^2+3y+5=0 ?

The solutions to the equation -y^2+3y+5=0 are y=-(-3+sqrt(29))/2 ,y=(3+sqrt(29))/2

Find the zeros of -y^2+3y+5=0

The zeros of -y^2+3y+5=0 are y=-(-3+sqrt(29))/2 ,y=(3+sqrt(29))/2

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

y = - \frac{- 3 + 29}{2}, y = \frac{3 + 29}{2}

Decimal Notation

y = - 1.19258 \dots, y = 4.19258 \dots

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 (- 1) \cdot 5 = 29

y_{1, 2} = \frac{- 3 \pm 29}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 3 + 29}{2 ( - 1 )}, y_{2} = \frac{- 3 - 29}{2 ( - 1 )}

y = \frac{- 3 + 29}{2 ( - 1 )} : - \frac{- 3 + 29}{2}

y = \frac{- 3 - 29}{2 ( - 1 )} : \frac{3 + 29}{2}

The solutions to the quadratic equation are:

y = - \frac{- 3 + 29}{2}, y = \frac{3 + 29}{2}

12 - (- 2)

0 - \frac{5}{2}

\frac{24 m ^{8} n ^{5}}{( - m ^{2} n ) ^{3} + 5 m ^{6} n ^{3}} \cdot (- 3 m n^{4})^{2}

3 x + 6 \leq 42

5 x^{2} - 2 x = 4

What are the solutions to the equation -y^2+3y+5=0 ?
The solutions to the equation -y^2+3y+5=0 are y=-(-3+sqrt(29))/2 ,y=(3+sqrt(29))/2
Find the zeros of -y^2+3y+5=0
The zeros of -y^2+3y+5=0 are y=-(-3+sqrt(29))/2 ,y=(3+sqrt(29))/2