What are the solutions to the equation z^2-7z-3=0 ?

The solutions to the equation z^2-7z-3=0 are z=(7+sqrt(61))/2 ,z=(7-sqrt(61))/2

Find the zeros of z^2-7z-3=0

The zeros of z^2-7z-3=0 are z=(7+sqrt(61))/2 ,z=(7-sqrt(61))/2

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z^{2} - 7 z - 3 = 0

z = \frac{7 + 61}{2}, z = \frac{7 - 61}{2}

Decimal

z = 7.40512 \dots, z = - 0.40512 \dots

Solution steps

z^{2} - 7 z - 3 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

(- 7)^{2} - 4 \cdot 1 \cdot (- 3) = 61

z_{1, 2} = \frac{- ( - 7 ) \pm 61}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 7 ) + 61}{2 \cdot 1}, z_{2} = \frac{- ( - 7 ) - 61}{2 \cdot 1}

z = \frac{- ( - 7 ) + 61}{2 \cdot 1} : \frac{7 + 61}{2}

z = \frac{- ( - 7 ) - 61}{2 \cdot 1} : \frac{7 - 61}{2}

The solutions to the quadratic equation are:

z = \frac{7 + 61}{2}, z = \frac{7 - 61}{2}

- 2 x^{2} - 10 x + 5 = 0

- a^{2} + a + 2 = 0

- 2 (x - 3)^{2} = 0

x^{2} + \frac{1}{2} x - 5 = 0

2 n^{2} - 5 n + 2 = 0

What are the solutions to the equation z^2-7z-3=0 ?
The solutions to the equation z^2-7z-3=0 are z=(7+sqrt(61))/2 ,z=(7-sqrt(61))/2
Find the zeros of z^2-7z-3=0
The zeros of z^2-7z-3=0 are z=(7+sqrt(61))/2 ,z=(7-sqrt(61))/2