What are the solutions to the equation (z(z+1))/2 =21 ?

The solutions to the equation (z(z+1))/2 =21 are z=6,z=-7

Find the zeros of (z(z+1))/2 =21

The zeros of (z(z+1))/2 =21 are z=6,z=-7

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\frac{z ( z + 1 )}{2} = 21

z = 6, z = - 7

Solution steps

\frac{z ( z + 1 )}{2} = 21

Multiply both sides by

2

z (z + 1) = 42

Expand

z (z + 1) : z^{2} + z

z^{2} + z = 42

Move

42

to the left side

z^{2} + z - 42 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

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

Separate the solutions

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

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

z = \frac{- 1 - 13}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

z = 6, z = - 7

a^{2} + 7 a + 6 = 0

7 x^{2} = 12 x

(7 z + 1)^{2} = - 1

z^{2} - 3 z - 7 = 0

(x - 4)^{2} - 9 = x

What are the solutions to the equation (z(z+1))/2 =21 ?
The solutions to the equation (z(z+1))/2 =21 are z=6,z=-7
Find the zeros of (z(z+1))/2 =21
The zeros of (z(z+1))/2 =21 are z=6,z=-7