What are the solutions to the equation z^2-7z+4=0 ?

The solutions to the equation z^2-7z+4=0 are z=(7+sqrt(33))/2 ,z=(7-sqrt(33))/2

Find the zeros of z^2-7z+4=0

The zeros of z^2-7z+4=0 are z=(7+sqrt(33))/2 ,z=(7-sqrt(33))/2

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z^{2} - 7 z + 4 = 0

z = \frac{7 + 33}{2}, z = \frac{7 - 33}{2}

Decimal Notation

z = 6.37228 \dots, z = 0.62771 \dots

Solution steps

z^{2} - 7 z + 4 = 0

Solve with the quadratic formula

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

(- 7)^{2} - 4 \cdot 1 \cdot 4 = 33

z_{1, 2} = \frac{- ( - 7 ) \pm 33}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 7 ) + 33}{2 \cdot 1}, z_{2} = \frac{- ( - 7 ) - 33}{2 \cdot 1}

z = \frac{- ( - 7 ) + 33}{2 \cdot 1} : \frac{7 + 33}{2}

z = \frac{- ( - 7 ) - 33}{2 \cdot 1} : \frac{7 - 33}{2}

The solutions to the quadratic equation are:

z = \frac{7 + 33}{2}, z = \frac{7 - 33}{2}

5 x^{2} + 23 x = 2 x - 4

x, x^{2} y^{2} = 2

r^{2} + 8 r + 784 = 0

4 x^{2} - 11 = 53

7 x + 450 = x^{2}

What are the solutions to the equation z^2-7z+4=0 ?
The solutions to the equation z^2-7z+4=0 are z=(7+sqrt(33))/2 ,z=(7-sqrt(33))/2
Find the zeros of z^2-7z+4=0
The zeros of z^2-7z+4=0 are z=(7+sqrt(33))/2 ,z=(7-sqrt(33))/2