What are the solutions to the equation r^2+13r+42=0 ?

The solutions to the equation r^2+13r+42=0 are r=-6,r=-7

Find the zeros of r^2+13r+42=0

The zeros of r^2+13r+42=0 are r=-6,r=-7

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r^{2} + 13 r + 42 = 0

r = - 6, r = - 7

Solution steps

r^{2} + 13 r + 42 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 42}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 42 = 1

r_{1, 2} = \frac{- 13 \pm 1}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 13 + 1}{2 \cdot 1}, r_{2} = \frac{- 13 - 1}{2 \cdot 1}

r = \frac{- 13 + 1}{2 \cdot 1} : - 6

r = \frac{- 13 - 1}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

r = - 6, r = - 7

x^{2} - 18 x = - 56

x^{2} - 23 x + 3 = 0

a, \frac{d ^{3}}{d t ^{3}} ((u) u^{3}) = a^{2} t

0 = - 2 x^{2} + x (3 - 6) - 8

2 n^{2} - 4 n + 2 = 0

What are the solutions to the equation r^2+13r+42=0 ?
The solutions to the equation r^2+13r+42=0 are r=-6,r=-7
Find the zeros of r^2+13r+42=0
The zeros of r^2+13r+42=0 are r=-6,r=-7