What are the solutions to the equation r^2-9r-38=-9 ?

The solutions to the equation r^2-9r-38=-9 are r=(9+sqrt(197))/2 ,r=(9-sqrt(197))/2

Find the zeros of r^2-9r-38=-9

The zeros of r^2-9r-38=-9 are r=(9+sqrt(197))/2 ,r=(9-sqrt(197))/2

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r^{2} - 9 r - 38 = - 9

r = \frac{9 + 197}{2}, r = \frac{9 - 197}{2}

Decimal Notation

r = 11.51783 \dots, r = - 2.51783 \dots

Solution steps

r^{2} - 9 r - 38 = - 9

Move

9

to the left side

r^{2} - 9 r - 29 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 1 \cdot ( - 29 )}{2 \cdot 1}

(- 9)^{2} - 4 \cdot 1 \cdot (- 29) = 197

r_{1, 2} = \frac{- ( - 9 ) \pm 197}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 9 ) + 197}{2 \cdot 1}, r_{2} = \frac{- ( - 9 ) - 197}{2 \cdot 1}

r = \frac{- ( - 9 ) + 197}{2 \cdot 1} : \frac{9 + 197}{2}

r = \frac{- ( - 9 ) - 197}{2 \cdot 1} : \frac{9 - 197}{2}

The solutions to the quadratic equation are:

r = \frac{9 + 197}{2}, r = \frac{9 - 197}{2}

3 x - 1 = 2 x + 2

x + 3 = x^{2} + 1

lo g_{8} (2) + lo g_{8} (32)

\frac{1}{6} - \frac{1}{5}

What are the solutions to the equation r^2-9r-38=-9 ?
The solutions to the equation r^2-9r-38=-9 are r=(9+sqrt(197))/2 ,r=(9-sqrt(197))/2
Find the zeros of r^2-9r-38=-9
The zeros of r^2-9r-38=-9 are r=(9+sqrt(197))/2 ,r=(9-sqrt(197))/2