What are the solutions to the equation t^2-4t=7 ?

The solutions to the equation t^2-4t=7 are t=2+sqrt(11),t=2-sqrt(11)

Find the zeros of t^2-4t=7

The zeros of t^2-4t=7 are t=2+sqrt(11),t=2-sqrt(11)

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t^{2} - 4 t = 7

t = 2 + 11, t = 2 - 11

Decimal Notation

t = 5.31662 \dots, t = - 1.31662 \dots

Solution steps

t^{2} - 4 t = 7

Move

7

to the left side

t^{2} - 4 t - 7 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot (- 7) = 211

t_{1, 2} = \frac{- ( - 4 ) \pm 2 11}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 4 ) + 2 11}{2 \cdot 1}, t_{2} = \frac{- ( - 4 ) - 2 11}{2 \cdot 1}

t = \frac{- ( - 4 ) + 2 11}{2 \cdot 1} : 2 + 11

t = \frac{- ( - 4 ) - 2 11}{2 \cdot 1} : 2 - 11

The solutions to the quadratic equation are:

t = 2 + 11, t = 2 - 11

- 2 = - 4 + x

42 < 2 m - 10

\frac{3 x + 3 y}{6}

4 \leq x - \frac{1}{2} \leq 9

8 w^{2} + w + 1 = 0

What are the solutions to the equation t^2-4t=7 ?
The solutions to the equation t^2-4t=7 are t=2+sqrt(11),t=2-sqrt(11)
Find the zeros of t^2-4t=7
The zeros of t^2-4t=7 are t=2+sqrt(11),t=2-sqrt(11)