What are the solutions to the equation t^2+4=4sqrt(2)t ?

The solutions to the equation t^2+4=4sqrt(2)t are t=2(1+sqrt(2)),t=2(sqrt(2)-1)

Find the zeros of t^2+4=4sqrt(2)t

The zeros of t^2+4=4sqrt(2)t are t=2(1+sqrt(2)),t=2(sqrt(2)-1)

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t^{2} + 4 = 42 t

t = 2 (1 + 2), t = 2 (2 - 1)

Decimal Notation

t = 4.82842 \dots, t = 0.82842 \dots

Solution steps

t^{2} + 4 = 42 t

Move

42 t

to the left side

t^{2} + 4 - 42 t = 0

Write in the standard form

a x^{2} + b x + c = 0

t^{2} - 42 t + 4 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

t = 2 (1 + 2), t = 2 (2 - 1)

4 n^{2} + 72 n - 76

3 (x + 9)^{2} = 54

(x + 3) (x + 8) = 0

3108

3 x^{4} y^{2} - 9 x^{2} y^{2}

What are the solutions to the equation t^2+4=4sqrt(2)t ?
The solutions to the equation t^2+4=4sqrt(2)t are t=2(1+sqrt(2)),t=2(sqrt(2)-1)
Find the zeros of t^2+4=4sqrt(2)t
The zeros of t^2+4=4sqrt(2)t are t=2(1+sqrt(2)),t=2(sqrt(2)-1)