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

The solutions to the equation 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1 are t=0

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

The zeros of 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1 are t=0

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

t = 0

Solution steps

1 = (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

1 = 4 t^{2} + 1

Switch sides

4 t^{2} + 1 = 1

Move

1

to the left side

4 t^{2} = 0

Solve with the quadratic formula

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

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

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

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

\frac{- 0}{2 \cdot 4} = 0

t = 0

The solution to the quadratic equation is:

t = 0

x^{2} - 3 x + 1 = 0

0 = 3 x^{2} - 2 x - 1

(x + 9) (x - 7) = 0

2 x^{2} + 22 x + 36 = 0

y^{2} + 8 y + 11 = 0

What are the solutions to the equation 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1 ?
The solutions to the equation 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1 are t=0
Find the zeros of 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1
The zeros of 1=(1-2(1)+5)t^2+(2(1)-2(1))t+1 are t=0