What are the solutions to the equation t^2-5t+10=0 ?

The solutions to the equation t^2-5t+10=0 are t= 5/2+i(sqrt(15))/2 ,t= 5/2-i(sqrt(15))/2

Find the zeros of t^2-5t+10=0

The zeros of t^2-5t+10=0 are t= 5/2+i(sqrt(15))/2 ,t= 5/2-i(sqrt(15))/2

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t^{2} - 5 t + 10 = 0

t = \frac{5}{2} + i \frac{15}{2}, t = \frac{5}{2} - i \frac{15}{2}

Solution steps

t^{2} - 5 t + 10 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 10}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 10 : 15 i

t_{1, 2} = \frac{- ( - 5 ) \pm 15 i}{2 \cdot 1}

Separate the solutions

t_{1} = \frac{- ( - 5 ) + 15 i}{2 \cdot 1}, t_{2} = \frac{- ( - 5 ) - 15 i}{2 \cdot 1}

t = \frac{- ( - 5 ) + 15 i}{2 \cdot 1} : \frac{5}{2} + i \frac{15}{2}

t = \frac{- ( - 5 ) - 15 i}{2 \cdot 1} : \frac{5}{2} - i \frac{15}{2}

The solutions to the quadratic equation are:

t = \frac{5}{2} + i \frac{15}{2}, t = \frac{5}{2} - i \frac{15}{2}

x^{2} - 81 > 0

16 - c^{2}

\frac{3}{7}

(x + 2) (x - 3) (x + 8) \leq 0

3 16 x^{4} y^{4} z

What are the solutions to the equation t^2-5t+10=0 ?
The solutions to the equation t^2-5t+10=0 are t= 5/2+i(sqrt(15))/2 ,t= 5/2-i(sqrt(15))/2
Find the zeros of t^2-5t+10=0
The zeros of t^2-5t+10=0 are t= 5/2+i(sqrt(15))/2 ,t= 5/2-i(sqrt(15))/2