What are the solutions to the equation t^2-t=12 ?

The solutions to the equation t^2-t=12 are t=4,t=-3

Find the zeros of t^2-t=12

The zeros of t^2-t=12 are t=4,t=-3

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t^{2} - t = 12

t = 4, t = - 3

Solution steps

t^{2} - t = 12

Move

12

to the left side

t^{2} - t - 12 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

t_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 1}, t_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 1}

t = \frac{- ( - 1 ) + 7}{2 \cdot 1} : 4

t = \frac{- ( - 1 ) - 7}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

t = 4, t = - 3

∣ x ∣ + 9 \leq 7

6 m^{3} n^{3} \cdot 8 m^{2} n^{3}

- 3 x < - 6

2 a^{2} - a

(\frac{n ^{8}}{n ^{2}})^{2}

What are the solutions to the equation t^2-t=12 ?
The solutions to the equation t^2-t=12 are t=4,t=-3
Find the zeros of t^2-t=12
The zeros of t^2-t=12 are t=4,t=-3