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

The solutions to the equation -z^2-7z-4=-2z^2 are z=(7+sqrt(65))/2 ,z=(7-sqrt(65))/2

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

The zeros of -z^2-7z-4=-2z^2 are z=(7+sqrt(65))/2 ,z=(7-sqrt(65))/2

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

z = \frac{7 + 65}{2}, z = \frac{7 - 65}{2}

Decimal Notation

z = 7.53112 \dots, z = - 0.53112 \dots

Solution steps

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

Move

2 z^{2}

to the left side

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) = 65

z_{1, 2} = \frac{- ( - 7 ) \pm 65}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- ( - 7 ) + 65}{2 \cdot 1}, z_{2} = \frac{- ( - 7 ) - 65}{2 \cdot 1}

z = \frac{- ( - 7 ) + 65}{2 \cdot 1} : \frac{7 + 65}{2}

z = \frac{- ( - 7 ) - 65}{2 \cdot 1} : \frac{7 - 65}{2}

The solutions to the quadratic equation are:

z = \frac{7 + 65}{2}, z = \frac{7 - 65}{2}

x, f (x - 2) = - 3 x^{2} - 1

x, y (x^{2} - 1) = 2

6 - 2 x^{2} = 6 x

x^{2} + 40 = - 14 x

x^{2} + 2 = 4 x

What are the solutions to the equation -z^2-7z-4=-2z^2 ?
The solutions to the equation -z^2-7z-4=-2z^2 are z=(7+sqrt(65))/2 ,z=(7-sqrt(65))/2
Find the zeros of -z^2-7z-4=-2z^2
The zeros of -z^2-7z-4=-2z^2 are z=(7+sqrt(65))/2 ,z=(7-sqrt(65))/2