What are the solutions to the equation 2t^2-6t+5=0 ?

The solutions to the equation 2t^2-6t+5=0 are t= 3/2+i 1/2 ,t= 3/2-i 1/2

Find the zeros of 2t^2-6t+5=0

The zeros of 2t^2-6t+5=0 are t= 3/2+i 1/2 ,t= 3/2-i 1/2

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2 t^{2} - 6 t + 5 = 0

t = \frac{3}{2} + i \frac{1}{2}, t = \frac{3}{2} - i \frac{1}{2}

Solution steps

2 t^{2} - 6 t + 5 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 2 \cdot 5}{2 \cdot 2}

Simplify

(- 6)^{2} - 4 \cdot 2 \cdot 5 : 2 i

t_{1, 2} = \frac{- ( - 6 ) \pm 2 i}{2 \cdot 2}

Separate the solutions

t_{1} = \frac{- ( - 6 ) + 2 i}{2 \cdot 2}, t_{2} = \frac{- ( - 6 ) - 2 i}{2 \cdot 2}

t = \frac{- ( - 6 ) + 2 i}{2 \cdot 2} : \frac{3}{2} + i \frac{1}{2}

t = \frac{- ( - 6 ) - 2 i}{2 \cdot 2} : \frac{3}{2} - i \frac{1}{2}

The solutions to the quadratic equation are:

t = \frac{3}{2} + i \frac{1}{2}, t = \frac{3}{2} - i \frac{1}{2}

- X \leq 0

\frac{2 a ^{2} b ( 3 a ^{3} c ^{5} )}{12 b ^{3} c ^{4}}

8 b^{2} - 22 b + 5

\frac{2 π}{3} \cdot \frac{180}{π}

∣ x + 1 ∣ + ∣ x - 3 ∣ = 6

What are the solutions to the equation 2t^2-6t+5=0 ?
The solutions to the equation 2t^2-6t+5=0 are t= 3/2+i 1/2 ,t= 3/2-i 1/2
Find the zeros of 2t^2-6t+5=0
The zeros of 2t^2-6t+5=0 are t= 3/2+i 1/2 ,t= 3/2-i 1/2