What are the solutions to the equation -9r+r^2+23=0 ?

The solutions to the equation -9r+r^2+23=0 are r= 9/2+i(sqrt(11))/2 ,r= 9/2-i(sqrt(11))/2

Find the zeros of -9r+r^2+23=0

The zeros of -9r+r^2+23=0 are r= 9/2+i(sqrt(11))/2 ,r= 9/2-i(sqrt(11))/2

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- 9 r + r^{2} + 23 = 0

r = \frac{9}{2} + i \frac{11}{2}, r = \frac{9}{2} - i \frac{11}{2}

Solution steps

- 9 r + r^{2} + 23 = 0

Write in the standard form

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

r^{2} - 9 r + 23 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot 23}{2 \cdot 1}

Simplify

(- 9)^{2} - 4 \cdot 1 \cdot 23 : 11 i

r_{1, 2} = \frac{- ( - 9 ) \pm 11 i}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 9 ) + 11 i}{2 \cdot 1}, r_{2} = \frac{- ( - 9 ) - 11 i}{2 \cdot 1}

r = \frac{- ( - 9 ) + 11 i}{2 \cdot 1} : \frac{9}{2} + i \frac{11}{2}

r = \frac{- ( - 9 ) - 11 i}{2 \cdot 1} : \frac{9}{2} - i \frac{11}{2}

The solutions to the quadratic equation are:

r = \frac{9}{2} + i \frac{11}{2}, r = \frac{9}{2} - i \frac{11}{2}

x + 2 = x^{2} - 8 x + 16

x, 4 x^{2} + 4 y^{2} + z^{2} = 64

x, f = 2 x^{2} + y^{2}

5 x^{2} + x + 10 = 0

X^{2} + 2 X + 5 = 0

What are the solutions to the equation -9r+r^2+23=0 ?
The solutions to the equation -9r+r^2+23=0 are r= 9/2+i(sqrt(11))/2 ,r= 9/2-i(sqrt(11))/2
Find the zeros of -9r+r^2+23=0
The zeros of -9r+r^2+23=0 are r= 9/2+i(sqrt(11))/2 ,r= 9/2-i(sqrt(11))/2