What are the solutions to the equation -t^2+10t=16 ?

The solutions to the equation -t^2+10t=16 are t=2,t=8

Find the zeros of -t^2+10t=16

The zeros of -t^2+10t=16 are t=2,t=8

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- t^{2} + 10 t = 16

t = 2, t = 8

Solution steps

- t^{2} + 10 t = 16

Move

16

to the left side

- t^{2} + 10 t - 16 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 ( - 1 ) ( - 16 )}{2 ( - 1 )}

1 0^{2} - 4 (- 1) (- 16) = 6

t_{1, 2} = \frac{- 10 \pm 6}{2 ( - 1 )}

Separate the solutions

t_{1} = \frac{- 10 + 6}{2 ( - 1 )}, t_{2} = \frac{- 10 - 6}{2 ( - 1 )}

t = \frac{- 10 + 6}{2 ( - 1 )} : 2

t = \frac{- 10 - 6}{2 ( - 1 )} : 8

The solutions to the quadratic equation are:

t = 2, t = 8

2 x^{2} - 11 x + 3 = 0

(18 - y)^{2} + y^{2} = 180

5 x^{2} - 9 = 0

y, (16 - x) y^{2} = x^{3}

w^{2} - 8 w + 15 = 0

What are the solutions to the equation -t^2+10t=16 ?
The solutions to the equation -t^2+10t=16 are t=2,t=8
Find the zeros of -t^2+10t=16
The zeros of -t^2+10t=16 are t=2,t=8