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

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

Find the zeros of x^2+13=4x

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

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x^{2} + 13 = 4 x

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

Solution steps

x^{2} + 13 = 4 x

Move

4 x

to the left side

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

Write in the standard form

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

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

i^{123}

\frac{3}{4} + 7

\frac{6 - i}{1 + i}

\frac{8 ^{5}}{8 ^{3}}

- 2 (x + 6) + 4 (x + 3) = 5 x - 2 (2 x + 2) + 4

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