What are the solutions to the equation r^2-3r-18=0 ?

The solutions to the equation r^2-3r-18=0 are r=6,r=-3

Find the zeros of r^2-3r-18=0

The zeros of r^2-3r-18=0 are r=6,r=-3

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r^{2} - 3 r - 18 = 0

r = 6, r = - 3

Solution steps

r^{2} - 3 r - 18 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 18 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 18) = 9

r_{1, 2} = \frac{- ( - 3 ) \pm 9}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 3 ) + 9}{2 \cdot 1}, r_{2} = \frac{- ( - 3 ) - 9}{2 \cdot 1}

r = \frac{- ( - 3 ) + 9}{2 \cdot 1} : 6

r = \frac{- ( - 3 ) - 9}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

r = 6, r = - 3

4 x^{2} - 7 x + 3

- 8 < - 4 x \leq 12

16 t^{2} - 25 s^{2}

\frac{4}{7} x + \frac{6}{7} = - \frac{2}{7}

- 2 x - 3 > 5

What are the solutions to the equation r^2-3r-18=0 ?
The solutions to the equation r^2-3r-18=0 are r=6,r=-3
Find the zeros of r^2-3r-18=0
The zeros of r^2-3r-18=0 are r=6,r=-3