What are the solutions to the equation (1+2t)^2=1+6t ?

The solutions to the equation (1+2t)^2=1+6t are t= 1/2 ,t=0

Find the zeros of (1+2t)^2=1+6t

The zeros of (1+2t)^2=1+6t are t= 1/2 ,t=0

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(1 + 2 t)^{2} = 1 + 6 t

t = \frac{1}{2}, t = 0

Decimal Notation

t = 0.5, t = 0

Solution steps

(1 + 2 t)^{2} = 1 + 6 t

Expand

(1 + 2 t)^{2} : 1 + 4 t + 4 t^{2}

1 + 4 t + 4 t^{2} = 1 + 6 t

Move

6 t

to the left side

4 t^{2} - 2 t + 1 = 1

Move

1

to the left side

4 t^{2} - 2 t = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 4 \cdot 0 = 2

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

Separate the solutions

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

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

t = \frac{- ( - 2 ) - 2}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

t = \frac{1}{2}, t = 0

10 b - 45 = 3 (b - 15)

\frac{x}{x + 3} + 1 < 0

8 x^{2} = 2 x + 3

\frac{4}{2 3}

\frac{1}{x + 2} + \frac{1}{x + 3} \leq 0

What are the solutions to the equation (1+2t)^2=1+6t ?
The solutions to the equation (1+2t)^2=1+6t are t= 1/2 ,t=0
Find the zeros of (1+2t)^2=1+6t
The zeros of (1+2t)^2=1+6t are t= 1/2 ,t=0