What are the solutions to the equation r^2-r+144=0 ?

The solutions to the equation r^2-r+144=0 are r= 1/2+i(5sqrt(23))/2 ,r= 1/2-i(5sqrt(23))/2

Find the zeros of r^2-r+144=0

The zeros of r^2-r+144=0 are r= 1/2+i(5sqrt(23))/2 ,r= 1/2-i(5sqrt(23))/2

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r^{2} - r + 144 = 0

r = \frac{1}{2} + i \frac{5 23}{2}, r = \frac{1}{2} - i \frac{5 23}{2}

Solution steps

r^{2} - r + 144 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot 144}{2 \cdot 1}

Simplify

(- 1)^{2} - 4 \cdot 1 \cdot 144 : 523 i

r_{1, 2} = \frac{- ( - 1 ) \pm 5 23 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 1 ) + 5 23 i}{2 \cdot 1}, r_{2} = \frac{- ( - 1 ) - 5 23 i}{2 \cdot 1}

r = \frac{- ( - 1 ) + 5 23 i}{2 \cdot 1} : \frac{1}{2} + i \frac{5 23}{2}

r = \frac{- ( - 1 ) - 5 23 i}{2 \cdot 1} : \frac{1}{2} - i \frac{5 23}{2}

The solutions to the quadratic equation are:

r = \frac{1}{2} + i \frac{5 23}{2}, r = \frac{1}{2} - i \frac{5 23}{2}

(\frac{1}{25})^{- \frac{3}{2}}

2 x^{2} + 8 x - 4 = 0

\frac{3}{4 ( \frac{1}{4} + 2 ) ^{\frac{5}{2}}}

10 x + 5 < - 8 + 6 x

\frac{1}{2 e ^{- 3}}

What are the solutions to the equation r^2-r+144=0 ?
The solutions to the equation r^2-r+144=0 are r= 1/2+i(5sqrt(23))/2 ,r= 1/2-i(5sqrt(23))/2
Find the zeros of r^2-r+144=0
The zeros of r^2-r+144=0 are r= 1/2+i(5sqrt(23))/2 ,r= 1/2-i(5sqrt(23))/2