What are the solutions to the equation r^2+6r-7=0 ?

The solutions to the equation r^2+6r-7=0 are r=1,r=-7

Find the zeros of r^2+6r-7=0

The zeros of r^2+6r-7=0 are r=1,r=-7

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r^{2} + 6 r - 7 = 0

r = 1, r = - 7

Solution steps

r^{2} + 6 r - 7 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

6^{2} - 4 \cdot 1 \cdot (- 7) = 8

r_{1, 2} = \frac{- 6 \pm 8}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 6 + 8}{2 \cdot 1}, r_{2} = \frac{- 6 - 8}{2 \cdot 1}

r = \frac{- 6 + 8}{2 \cdot 1} : 1

r = \frac{- 6 - 8}{2 \cdot 1} : - 7

The solutions to the quadratic equation are:

r = 1, r = - 7

- 10 = - 2 + \frac{1}{2} d

\frac{3}{2} x - 1 = 5

5 x^{2} - 19 x = 4

ln (2) + ln (5)

i^{1}

What are the solutions to the equation r^2+6r-7=0 ?
The solutions to the equation r^2+6r-7=0 are r=1,r=-7
Find the zeros of r^2+6r-7=0
The zeros of r^2+6r-7=0 are r=1,r=-7