What are the solutions to the equation 6=1+20t-5t^2 ?

The solutions to the equation 6=1+20t-5t^2 are t=2-sqrt(3),t=2+sqrt(3)

Find the zeros of 6=1+20t-5t^2

The zeros of 6=1+20t-5t^2 are t=2-sqrt(3),t=2+sqrt(3)

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6 = 1 + 20 t - 5 t^{2}

t = 2 - 3, t = 2 + 3

Decimal Notation

t = 0.26794 \dots, t = 3.73205 \dots

Solution steps

6 = 1 + 20 t - 5 t^{2}

Switch sides

1 + 20 t - 5 t^{2} = 6

Move

6

to the left side

- 5 t^{2} + 20 t - 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 5 ) ( - 5 )}{2 ( - 5 )}

2 0^{2} - 4 (- 5) (- 5) = 103

t_{1, 2} = \frac{- 20 \pm 10 3}{2 ( - 5 )}

Separate the solutions

t_{1} = \frac{- 20 + 10 3}{2 ( - 5 )}, t_{2} = \frac{- 20 - 10 3}{2 ( - 5 )}

t = \frac{- 20 + 10 3}{2 ( - 5 )} : 2 - 3

t = \frac{- 20 - 10 3}{2 ( - 5 )} : 2 + 3

The solutions to the quadratic equation are:

t = 2 - 3, t = 2 + 3

5^{2} + 1 2^{2} = c^{2}

x^{2} = 4 x + 3

\frac{5 j ^{4}}{( 2 j ^{4} ) ^{4}}

144 = - 12 (x + 5)

2 ∣ x - 2 ∣ > 8

What are the solutions to the equation 6=1+20t-5t^2 ?
The solutions to the equation 6=1+20t-5t^2 are t=2-sqrt(3),t=2+sqrt(3)
Find the zeros of 6=1+20t-5t^2
The zeros of 6=1+20t-5t^2 are t=2-sqrt(3),t=2+sqrt(3)