What are the solutions to the equation -y^2-5y+7=0 ?

The solutions to the equation -y^2-5y+7=0 are y=-(5+sqrt(53))/2 ,y=(sqrt(53)-5)/2

Find the zeros of -y^2-5y+7=0

The zeros of -y^2-5y+7=0 are y=-(5+sqrt(53))/2 ,y=(sqrt(53)-5)/2

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- y^{2} - 5 y + 7 = 0

y = - \frac{5 + 53}{2}, y = \frac{53 - 5}{2}

Decimal Notation

y = - 6.14005 \dots, y = 1.14005 \dots

Solution steps

- y^{2} - 5 y + 7 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( - 1 ) \cdot 7}{2 ( - 1 )}

(- 5)^{2} - 4 (- 1) \cdot 7 = 53

y_{1, 2} = \frac{- ( - 5 ) \pm 53}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 5 ) + 53}{2 ( - 1 )}, y_{2} = \frac{- ( - 5 ) - 53}{2 ( - 1 )}

y = \frac{- ( - 5 ) + 53}{2 ( - 1 )} : - \frac{5 + 53}{2}

y = \frac{- ( - 5 ) - 53}{2 ( - 1 )} : \frac{53 - 5}{2}

The solutions to the quadratic equation are:

y = - \frac{5 + 53}{2}, y = \frac{53 - 5}{2}

2 x^{2} + 5 x - 25 = 0

\frac{3}{6} - \frac{5}{10}

A^{2} (\frac{π}{\frac{1}{2}}) = 1

\frac{4 x + 1}{3} = \frac{x + 2}{4}

\frac{5}{6} x + \frac{1}{5} \leq \frac{1}{30}

What are the solutions to the equation -y^2-5y+7=0 ?
The solutions to the equation -y^2-5y+7=0 are y=-(5+sqrt(53))/2 ,y=(sqrt(53)-5)/2
Find the zeros of -y^2-5y+7=0
The zeros of -y^2-5y+7=0 are y=-(5+sqrt(53))/2 ,y=(sqrt(53)-5)/2