What are the solutions to the equation r^2-2r-15=0 ?

The solutions to the equation r^2-2r-15=0 are r=5,r=-3

Find the zeros of r^2-2r-15=0

The zeros of r^2-2r-15=0 are r=5,r=-3

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r^{2} - 2 r - 15 = 0

r = 5, r = - 3

Solution steps

r^{2} - 2 r - 15 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 15) = 8

r_{1, 2} = \frac{- ( - 2 ) \pm 8}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 1}, r_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 1}

r = \frac{- ( - 2 ) + 8}{2 \cdot 1} : 5

r = \frac{- ( - 2 ) - 8}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

r = 5, r = - 3

- 2 x^{2} - 8 x - 14 = - 6

2 x^{2} - 7 x = - 6

25 x^{2} - 10 x + 4 = 0

y, x^{2} + x y + y^{2} = 1

x^{2} + 6 x + 9 = 0

What are the solutions to the equation r^2-2r-15=0 ?
The solutions to the equation r^2-2r-15=0 are r=5,r=-3
Find the zeros of r^2-2r-15=0
The zeros of r^2-2r-15=0 are r=5,r=-3