What are the solutions to the equation -5z^2-3z-11=-6z^2 ?

The solutions to the equation -5z^2-3z-11=-6z^2 are z=(3+sqrt(53))/2 ,z=(3-sqrt(53))/2

Find the zeros of -5z^2-3z-11=-6z^2

The zeros of -5z^2-3z-11=-6z^2 are z=(3+sqrt(53))/2 ,z=(3-sqrt(53))/2

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

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

Solution steps

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

Move

6 z^{2}

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}

x^{2} = 128

x + 7 = 17

9 x - 8 x = 107

8 (x + 2) - 3 (2 x + 1) = 2 (x + 5)

\frac{y - \frac{x ^{2}}{y}}{y ^{2}}

What are the solutions to the equation -5z^2-3z-11=-6z^2 ?
The solutions to the equation -5z^2-3z-11=-6z^2 are z=(3+sqrt(53))/2 ,z=(3-sqrt(53))/2
Find the zeros of -5z^2-3z-11=-6z^2
The zeros of -5z^2-3z-11=-6z^2 are z=(3+sqrt(53))/2 ,z=(3-sqrt(53))/2