What are the solutions to the equation x^2-5x+36=0 ?

The solutions to the equation x^2-5x+36=0 are x= 5/2+i(sqrt(119))/2 ,x= 5/2-i(sqrt(119))/2

Find the zeros of x^2-5x+36=0

The zeros of x^2-5x+36=0 are x= 5/2+i(sqrt(119))/2 ,x= 5/2-i(sqrt(119))/2

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x^{2} - 5 x + 36 = 0

x = \frac{5}{2} + i \frac{119}{2}, x = \frac{5}{2} - i \frac{119}{2}

Solution steps

x^{2} - 5 x + 36 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 1 \cdot 36}{2 \cdot 1}

Simplify

(- 5)^{2} - 4 \cdot 1 \cdot 36 : 119 i

x_{1, 2} = \frac{- ( - 5 ) \pm 119 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 5 ) + 119 i}{2 \cdot 1}, x_{2} = \frac{- ( - 5 ) - 119 i}{2 \cdot 1}

x = \frac{- ( - 5 ) + 119 i}{2 \cdot 1} : \frac{5}{2} + i \frac{119}{2}

x = \frac{- ( - 5 ) - 119 i}{2 \cdot 1} : \frac{5}{2} - i \frac{119}{2}

The solutions to the quadratic equation are:

x = \frac{5}{2} + i \frac{119}{2}, x = \frac{5}{2} - i \frac{119}{2}

16 (5 x - 6)^{2} - 41 = - 37

(x - 3) (x - 1) = 15

4 n^{2} - 9 n + 5 = 0

16 x^{2} = 12 x

x (5 x - 4) = 2

What are the solutions to the equation x^2-5x+36=0 ?
The solutions to the equation x^2-5x+36=0 are x= 5/2+i(sqrt(119))/2 ,x= 5/2-i(sqrt(119))/2
Find the zeros of x^2-5x+36=0
The zeros of x^2-5x+36=0 are x= 5/2+i(sqrt(119))/2 ,x= 5/2-i(sqrt(119))/2