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

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

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

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

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

r = 7, r = 6

Solution steps

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

Solve with the quadratic formula

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

(- 13)^{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} : 7

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

The solutions to the quadratic equation are:

r = 7, r = 6

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