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

The solutions to the equation 3z^2-z+5=6 are z=(1+sqrt(13))/6 ,z=(1-sqrt(13))/6

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

The zeros of 3z^2-z+5=6 are z=(1+sqrt(13))/6 ,z=(1-sqrt(13))/6

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

z = \frac{1 + 13}{6}, z = \frac{1 - 13}{6}

Decimal Notation

z = 0.76759 \dots, z = - 0.43425 \dots

Solution steps

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

Move

6

to the left side

3 z^{2} - z - 1 = 0

Solve with the quadratic formula

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

(- 1)^{2} - 4 \cdot 3 (- 1) = 13

z_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 \cdot 3}

Separate the solutions

z_{1} = \frac{- ( - 1 ) + 13}{2 \cdot 3}, z_{2} = \frac{- ( - 1 ) - 13}{2 \cdot 3}

z = \frac{- ( - 1 ) + 13}{2 \cdot 3} : \frac{1 + 13}{6}

z = \frac{- ( - 1 ) - 13}{2 \cdot 3} : \frac{1 - 13}{6}

The solutions to the quadratic equation are:

z = \frac{1 + 13}{6}, z = \frac{1 - 13}{6}

- 2 p^{2} + 13 p - 12 = - 6 p^{2}

x^{2} + 14 = 50

a, 3 a^{2} + 3 a + a t = \frac{30}{t}

7 A^{2} - 11 A - 6 = 0

3 (2 x + 8)^{2} + 18 = 45

What are the solutions to the equation 3z^2-z+5=6 ?
The solutions to the equation 3z^2-z+5=6 are z=(1+sqrt(13))/6 ,z=(1-sqrt(13))/6
Find the zeros of 3z^2-z+5=6
The zeros of 3z^2-z+5=6 are z=(1+sqrt(13))/6 ,z=(1-sqrt(13))/6