What are the solutions to the equation y^2+5y-4=0 ?

The solutions to the equation y^2+5y-4=0 are y=(-5+sqrt(41))/2 ,y=(-5-sqrt(41))/2

Find the zeros of y^2+5y-4=0

The zeros of y^2+5y-4=0 are y=(-5+sqrt(41))/2 ,y=(-5-sqrt(41))/2

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y^{2} + 5 y - 4 = 0

y = \frac{- 5 + 41}{2}, y = \frac{- 5 - 41}{2}

Decimal Notation

y = 0.70156 \dots, y = - 5.70156 \dots

Solution steps

y^{2} + 5 y - 4 = 0

Solve with the quadratic formula

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

5^{2} - 4 \cdot 1 \cdot (- 4) = 41

y_{1, 2} = \frac{- 5 \pm 41}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 5 + 41}{2 \cdot 1}, y_{2} = \frac{- 5 - 41}{2 \cdot 1}

y = \frac{- 5 + 41}{2 \cdot 1} : \frac{- 5 + 41}{2}

y = \frac{- 5 - 41}{2 \cdot 1} : \frac{- 5 - 41}{2}

The solutions to the quadratic equation are:

y = \frac{- 5 + 41}{2}, y = \frac{- 5 - 41}{2}

3 x^{2} + 21 x + 36 = 0

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6 y^{3} - 12 y^{4} x - 18 y^{5}

\frac{2 x}{7} + \frac{2}{3} = \frac{x}{21} + 2

\frac{10 x ^{4}}{y ^{- 1}}

What are the solutions to the equation y^2+5y-4=0 ?
The solutions to the equation y^2+5y-4=0 are y=(-5+sqrt(41))/2 ,y=(-5-sqrt(41))/2
Find the zeros of y^2+5y-4=0
The zeros of y^2+5y-4=0 are y=(-5+sqrt(41))/2 ,y=(-5-sqrt(41))/2