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

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

Find the zeros of z^2-3z-5=0

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

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

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

Decimal Notation

z = 4.19258 \dots, z = - 1.19258 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

\frac{b ^{2}}{b - 3} + \frac{9}{3 - b}

x^{2} - 15 x = - 56

8 (\frac{3}{4})

- 4 (v + 2) + 2 = 2 (v + 6)

5 a + 5

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