What are the solutions to the equation x^2-5x+24=0 ?

The solutions to the equation x^2-5x+24=0 are x= 5/2+i(sqrt(71))/2 ,x= 5/2-i(sqrt(71))/2

Find the zeros of x^2-5x+24=0

The zeros of x^2-5x+24=0 are x= 5/2+i(sqrt(71))/2 ,x= 5/2-i(sqrt(71))/2

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x^{2} - 5 x + 24 = 0

x = \frac{5}{2} + i \frac{71}{2}, x = \frac{5}{2} - i \frac{71}{2}

Solution steps

x^{2} - 5 x + 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 24}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 24 : 71 i

x_{1, 2} = \frac{- ( - 5 ) \pm 71 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 71 i}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 71 i}{2 \cdot 1}

x = \frac{- ( - 5 ) + 71 i}{2 \cdot 1} : \frac{5}{2} + i \frac{71}{2}

x = \frac{- ( - 5 ) - 71 i}{2 \cdot 1} : \frac{5}{2} - i \frac{71}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{71}{2}, x = \frac{5}{2} - i \frac{71}{2}

x^{2} - 12 x + 20

4 \cdot 9

\frac{1}{2} + \frac{1}{3} - \frac{1}{6}

3 x + 3 = 2 (x + 7)

1 + 5 ∣ 4 x - 10 ∣ = 46

What are the solutions to the equation x^2-5x+24=0 ?
The solutions to the equation x^2-5x+24=0 are x= 5/2+i(sqrt(71))/2 ,x= 5/2-i(sqrt(71))/2
Find the zeros of x^2-5x+24=0
The zeros of x^2-5x+24=0 are x= 5/2+i(sqrt(71))/2 ,x= 5/2-i(sqrt(71))/2