What are the solutions to the equation 21=(t^2-t)/2 ?

The solutions to the equation 21=(t^2-t)/2 are t=7,t=-6

Find the zeros of 21=(t^2-t)/2

The zeros of 21=(t^2-t)/2 are t=7,t=-6

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21 = \frac{t ^{2} - t}{2}

t = 7, t = - 6

Solution steps

21 = \frac{t ^{2} - t}{2}

Multiply both sides by

2

42 = t^{2} - t

Switch sides

t^{2} - t = 42

Move

42

to the left side

t^{2} - t - 42 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

t = \frac{- ( - 1 ) + 13}{2 \cdot 1} : 7

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

The solutions to the quadratic equation are:

t = 7, t = - 6

(13 n^{2} + 11 n - 2 n^{4}) + (- 13 n^{2} - 3 n - 6 n^{4})

\frac{1}{9 ^{8}}

(x - 3 i) (x + 3 i)

v^{11}

\frac{1}{12} + \frac{1}{24}

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