What are the solutions to the equation 55=(t^2-t)/2 ?

The solutions to the equation 55=(t^2-t)/2 are t=11,t=-10

Find the zeros of 55=(t^2-t)/2

The zeros of 55=(t^2-t)/2 are t=11,t=-10

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

t = 11, t = - 10

Solution steps

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

Multiply both sides by

2

110 = t^{2} - t

Switch sides

t^{2} - t = 110

Move

110

to the left side

t^{2} - t - 110 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 110 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 110) = 21

t_{1, 2} = \frac{- ( - 1 ) \pm 21}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 21}{2 \cdot 1}, t_{2} = \frac{- ( - 1 ) - 21}{2 \cdot 1}

t = \frac{- ( - 1 ) + 21}{2 \cdot 1} : 11

t = \frac{- ( - 1 ) - 21}{2 \cdot 1} : - 10

The solutions to the quadratic equation are:

t = 11, t = - 10

s + 9 \geq - 25

12 x + 20 > 4 (2 x + 2)

x + 2 > x

3 x^{2} + 8 x + 5

\frac{4 x + 5}{x - 8} \leq 0

What are the solutions to the equation 55=(t^2-t)/2 ?
The solutions to the equation 55=(t^2-t)/2 are t=11,t=-10
Find the zeros of 55=(t^2-t)/2
The zeros of 55=(t^2-t)/2 are t=11,t=-10