What are the solutions to the equation s^2-4s+13=0 ?

The solutions to the equation s^2-4s+13=0 are s=2+3i,s=2-3i

Find the zeros of s^2-4s+13=0

The zeros of s^2-4s+13=0 are s=2+3i,s=2-3i

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s^{2} - 4 s + 13 = 0

s = 2 + 3 i, s = 2 - 3 i

Solution steps

s^{2} - 4 s + 13 = 0

Solve with the quadratic formula

s_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 13}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 13 : 6 i

s_{1, 2} = \frac{- ( - 4 ) \pm 6 i}{2 \cdot 1}

Separate the solutions

s_{1} = \frac{- ( - 4 ) + 6 i}{2 \cdot 1}, s_{2} = \frac{- ( - 4 ) - 6 i}{2 \cdot 1}

s = \frac{- ( - 4 ) + 6 i}{2 \cdot 1} : 2 + 3 i

s = \frac{- ( - 4 ) - 6 i}{2 \cdot 1} : 2 - 3 i

The solutions to the quadratic equation are:

s = 2 + 3 i, s = 2 - 3 i

x (x - 3) = 5 (x - 2) + 1

2 x^{2} + 7 x + 7 = 0

9 y^{2} + 6 y + 1 = 0

x^{2} + 7 x + 12

x^{2} + (m - 3) x + m = 0

What are the solutions to the equation s^2-4s+13=0 ?
The solutions to the equation s^2-4s+13=0 are s=2+3i,s=2-3i
Find the zeros of s^2-4s+13=0
The zeros of s^2-4s+13=0 are s=2+3i,s=2-3i