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 = 4.19258 \dots, y = - 1.19258 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

f (y) = 0.5 y - 1 - (2 y + 4)

(x + 3)^{2} = - 4

15 x^{3} y^{5} - 25 x^{4} y^{2} + 10 x^{6} y^{4}

(1 + \frac{3 k}{n})^{3}

16 a^{3} - 2 b^{3}

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