What are the solutions to the equation t^2-2t=7 ?

The solutions to the equation t^2-2t=7 are t=1+2sqrt(2),t=1-2sqrt(2)

Find the zeros of t^2-2t=7

The zeros of t^2-2t=7 are t=1+2sqrt(2),t=1-2sqrt(2)

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t^{2} - 2 t = 7

t = 1 + 22, t = 1 - 22

Decimal Notation

t = 3.82842 \dots, t = - 1.82842 \dots

Solution steps

t^{2} - 2 t = 7

Move

7

to the left side

t^{2} - 2 t - 7 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 7) = 42

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

Separate the solutions

t_{1} = \frac{- ( - 2 ) + 4 2}{2 \cdot 1}, t_{2} = \frac{- ( - 2 ) - 4 2}{2 \cdot 1}

t = \frac{- ( - 2 ) + 4 2}{2 \cdot 1} : 1 + 22

t = \frac{- ( - 2 ) - 4 2}{2 \cdot 1} : 1 - 22

The solutions to the quadratic equation are:

t = 1 + 22, t = 1 - 22

\frac{4.1}{4.1 - 4}

500 \leq 12 x

- 54 x^{2} + 672 x = 0

2 x^{2} + 3 x - 5 > 0

\frac{16 2}{2 12}

What are the solutions to the equation t^2-2t=7 ?
The solutions to the equation t^2-2t=7 are t=1+2sqrt(2),t=1-2sqrt(2)
Find the zeros of t^2-2t=7
The zeros of t^2-2t=7 are t=1+2sqrt(2),t=1-2sqrt(2)