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

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

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

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

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

t = \frac{2 - 3}{2}, t = \frac{2 + 3}{2}

Decimal Notation

t = 0.13397 \dots, t = 1.86602 \dots

Solution steps

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

Expand

(1 - 2 (5) + 5) t^{2} + (2 (5) - 2 (1)) t + 1 : - 4 t^{2} + 8 t + 1

2 = - 4 t^{2} + 8 t + 1

Switch sides

- 4 t^{2} + 8 t + 1 = 2

Move

2

to the left side

- 4 t^{2} + 8 t - 1 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 1 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 1) = 43

t_{1, 2} = \frac{- 8 \pm 4 3}{2 ( - 4 )}

Separate the solutions

t_{1} = \frac{- 8 + 4 3}{2 ( - 4 )}, t_{2} = \frac{- 8 - 4 3}{2 ( - 4 )}

t = \frac{- 8 + 4 3}{2 ( - 4 )} : \frac{2 - 3}{2}

t = \frac{- 8 - 4 3}{2 ( - 4 )} : \frac{2 + 3}{2}

The solutions to the quadratic equation are:

t = \frac{2 - 3}{2}, t = \frac{2 + 3}{2}

m^{2} - 64 = 0

2 x^{2} - 15 = x

2 x^{2} + 20 x + 32 = 0

3 b^{2} - 19 b + 1 = - 5

x^{2} - 8 x + 7

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