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

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

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

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

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

x = 2 + 3 i, x = 2 - 3 i

Solution steps

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

Solve with the quadratic formula

x_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = 2 + 3 i, x = 2 - 3 i

z^{2} - 2 z - 35 = 0

3.23 = m (m - 5)

x^{2} + 9 x = 3

A (X) = X (\frac{288 - 2 X}{2})

4 x - 6 x^{2} = - 10

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