What are the solutions to the equation solvefor r,r^2+r-42=0 ?

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

Find the zeros of solvefor r,r^2+r-42=0

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

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r, r^{2} + r - 42 = 0

r = 6, r = - 7

Solution steps

r^{2} + r - 42 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot ( - 42 )}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot (- 42) = 13

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

r = 6, r = - 7

x, x^{2} - 12 x + 27 = 0

x^{2} - 9 x - 10 = 2 x + 10

x^{2} - 14 x + 30 = 0

3 x^{2} - 12 x + 8 = 0

- x^{2} + 21 x = 0

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