What are the solutions to the equation z^2-3z-11=-6z ?

The solutions to the equation z^2-3z-11=-6z are z=(-3+sqrt(53))/2 ,z=(-3-sqrt(53))/2

Find the zeros of z^2-3z-11=-6z

The zeros of z^2-3z-11=-6z are z=(-3+sqrt(53))/2 ,z=(-3-sqrt(53))/2

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z^{2} - 3 z - 11 = - 6 z

z = \frac{- 3 + 53}{2}, z = \frac{- 3 - 53}{2}

Decimal Notation

z = 2.14005 \dots, z = - 5.14005 \dots

Solution steps

z^{2} - 3 z - 11 = - 6 z

Move

6 z

to the left side

z^{2} + 3 z - 11 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 11 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 11) = 53

z_{1, 2} = \frac{- 3 \pm 53}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 3 + 53}{2 \cdot 1}, z_{2} = \frac{- 3 - 53}{2 \cdot 1}

z = \frac{- 3 + 53}{2 \cdot 1} : \frac{- 3 + 53}{2}

z = \frac{- 3 - 53}{2 \cdot 1} : \frac{- 3 - 53}{2}

The solutions to the quadratic equation are:

z = \frac{- 3 + 53}{2}, z = \frac{- 3 - 53}{2}

- 75 x^{2} - 15 x - 100 = - 150 x^{2} + 10 x + 250

- 2 x^{2} - x - 6 = 0

x, - 3^{2} + x^{2} = 1 0^{2}

s^{2} + 6 s + 144

5 x^{2} = - 9 x - 2

What are the solutions to the equation z^2-3z-11=-6z ?
The solutions to the equation z^2-3z-11=-6z are z=(-3+sqrt(53))/2 ,z=(-3-sqrt(53))/2
Find the zeros of z^2-3z-11=-6z
The zeros of z^2-3z-11=-6z are z=(-3+sqrt(53))/2 ,z=(-3-sqrt(53))/2