What are the solutions to the equation 4r^2-3r+9=4 ?

The solutions to the equation 4r^2-3r+9=4 are r= 3/8+i(sqrt(71))/8 ,r= 3/8-i(sqrt(71))/8

Find the zeros of 4r^2-3r+9=4

The zeros of 4r^2-3r+9=4 are r= 3/8+i(sqrt(71))/8 ,r= 3/8-i(sqrt(71))/8

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4 r^{2} - 3 r + 9 = 4

r = \frac{3}{8} + i \frac{71}{8}, r = \frac{3}{8} - i \frac{71}{8}

Solution steps

4 r^{2} - 3 r + 9 = 4

Move

4

to the left side

4 r^{2} - 3 r + 5 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 4 \cdot 5}{2 \cdot 4}

Simplify

(- 3)^{2} - 4 \cdot 4 \cdot 5 : 71 i

r_{1, 2} = \frac{- ( - 3 ) \pm 71 i}{2 \cdot 4}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 71 i}{2 \cdot 4}, r_{2} = \frac{- ( - 3 ) - 71 i}{2 \cdot 4}

r = \frac{- ( - 3 ) + 71 i}{2 \cdot 4} : \frac{3}{8} + i \frac{71}{8}

r = \frac{- ( - 3 ) - 71 i}{2 \cdot 4} : \frac{3}{8} - i \frac{71}{8}

The solutions to the quadratic equation are:

r = \frac{3}{8} + i \frac{71}{8}, r = \frac{3}{8} - i \frac{71}{8}

lo g_{9} (\frac{1}{81}) = x

ln (\frac{x}{y})

2 x + 4 < x - 1

\frac{x}{y} - 1

x^{2} + 26 x + 169 = 0

What are the solutions to the equation 4r^2-3r+9=4 ?
The solutions to the equation 4r^2-3r+9=4 are r= 3/8+i(sqrt(71))/8 ,r= 3/8-i(sqrt(71))/8
Find the zeros of 4r^2-3r+9=4
The zeros of 4r^2-3r+9=4 are r= 3/8+i(sqrt(71))/8 ,r= 3/8-i(sqrt(71))/8