What are the solutions to the equation -10t^2-12t=-9982 ?

The solutions to the equation -10t^2-12t=-9982 are t=-161/5 ,t=31

Find the zeros of -10t^2-12t=-9982

The zeros of -10t^2-12t=-9982 are t=-161/5 ,t=31

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- 10 t^{2} - 12 t = - 9982

t = - \frac{161}{5}, t = 31

Decimal Notation

t = - 32.2, t = 31

Solution steps

- 10 t^{2} - 12 t = - 9982

Move

9982

to the left side

- 10 t^{2} - 12 t + 9982 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 ( - 10 ) \cdot 9982}{2 ( - 10 )}

(- 12)^{2} - 4 (- 10) \cdot 9982 = 632

t_{1, 2} = \frac{- ( - 12 ) \pm 632}{2 ( - 10 )}

Separate the solutions

t_{1} = \frac{- ( - 12 ) + 632}{2 ( - 10 )}, t_{2} = \frac{- ( - 12 ) - 632}{2 ( - 10 )}

t = \frac{- ( - 12 ) + 632}{2 ( - 10 )} : - \frac{161}{5}

t = \frac{- ( - 12 ) - 632}{2 ( - 10 )} : 31

The solutions to the quadratic equation are:

t = - \frac{161}{5}, t = 31

3 x^{2} + 12 x - 4

3 x^{2} + 12 x = 63

\frac{5 x - 3}{3} + \frac{23 x + 25 x ^{2}}{15} = \frac{1 - 5 x ^{2} + 5 x}{5}

2 x^{2} - 12 x - 182 = 0

(3 x - 5)^{2} - (2 x + 1)^{2} = 24

What are the solutions to the equation -10t^2-12t=-9982 ?
The solutions to the equation -10t^2-12t=-9982 are t=-161/5 ,t=31
Find the zeros of -10t^2-12t=-9982
The zeros of -10t^2-12t=-9982 are t=-161/5 ,t=31