What are the solutions to the equation -1/2 (t-3)+t^2=0 ?

The solutions to the equation -1/2 (t-3)+t^2=0 are t= 1/4+i(sqrt(23))/4 ,t= 1/4-i(sqrt(23))/4

Find the zeros of -1/2 (t-3)+t^2=0

The zeros of -1/2 (t-3)+t^2=0 are t= 1/4+i(sqrt(23))/4 ,t= 1/4-i(sqrt(23))/4

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- \frac{1}{2} (t - 3) + t^{2} = 0

t = \frac{1}{4} + i \frac{23}{4}, t = \frac{1}{4} - i \frac{23}{4}

Solution steps

- \frac{1}{2} (t - 3) + t^{2} = 0

Multiply both sides by

2

- (t - 3) + 2 t^{2} = 0

Expand

- (t - 3) + 2 t^{2} : - t + 3 + 2 t^{2}

- t + 3 + 2 t^{2} = 0

Write in the standard form

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

2 t^{2} - t + 3 = 0

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 2 \cdot 3 : 23 i

t_{1, 2} = \frac{- ( - 1 ) \pm 23 i}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 1 ) + 23 i}{2 \cdot 2}, t_{2} = \frac{- ( - 1 ) - 23 i}{2 \cdot 2}

t = \frac{- ( - 1 ) + 23 i}{2 \cdot 2} : \frac{1}{4} + i \frac{23}{4}

t = \frac{- ( - 1 ) - 23 i}{2 \cdot 2} : \frac{1}{4} - i \frac{23}{4}

The solutions to the quadratic equation are:

t = \frac{1}{4} + i \frac{23}{4}, t = \frac{1}{4} - i \frac{23}{4}

- 0.3 x = 9

∣ 4 x + 2 ∣ - 3 = 7

0.625 x = 500

\frac{1}{4} (x + 5)^{2} - 1 = 3

x^{2} - 3 x = 0

What are the solutions to the equation -1/2 (t-3)+t^2=0 ?
The solutions to the equation -1/2 (t-3)+t^2=0 are t= 1/4+i(sqrt(23))/4 ,t= 1/4-i(sqrt(23))/4
Find the zeros of -1/2 (t-3)+t^2=0
The zeros of -1/2 (t-3)+t^2=0 are t= 1/4+i(sqrt(23))/4 ,t= 1/4-i(sqrt(23))/4