What are the solutions to the equation 3=(1-2(1)+5)t^2+(2(1)-2(1))t+1 ?

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

Find the zeros of 3=(1-2(1)+5)t^2+(2(1)-2(1))t+1

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

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3 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

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

Decimal Notation

t = 0.70710 \dots, t = - 0.70710 \dots

Solution steps

3 = (1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1

Expand

(1 - 2 (1) + 5) t^{2} + (2 (1) - 2 (1)) t + 1 : 4 t^{2} + 1

3 = 4 t^{2} + 1

Switch sides

4 t^{2} + 1 = 3

Move

3

to the left side

4 t^{2} - 2 = 0

Solve with the quadratic formula

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

0^{2} - 4 \cdot 4 (- 2) = 42

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

Separate the solutions

t_{1} = \frac{- 0 + 4 2}{2 \cdot 4}, t_{2} = \frac{- 0 - 4 2}{2 \cdot 4}

t = \frac{- 0 + 4 2}{2 \cdot 4} : \frac{1}{2}

t = \frac{- 0 - 4 2}{2 \cdot 4} : - \frac{1}{2}

The solutions to the quadratic equation are:

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

x^{2} + 4^{2} = 7^{2}

2 y^{2} + 16 y = - 8

(x - 4)^{2} + (x - 7)^{2} = 15

y, y^{2} (2 - x) = x^{3}

4 x^{2} + 44 x + 121 = 0

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