What are the solutions to the equation r^2-8r=-8 ?

The solutions to the equation r^2-8r=-8 are r=2(2+sqrt(2)),r=2(2-sqrt(2))

Find the zeros of r^2-8r=-8

The zeros of r^2-8r=-8 are r=2(2+sqrt(2)),r=2(2-sqrt(2))

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r^{2} - 8 r = - 8

r = 2 (2 + 2), r = 2 (2 - 2)

Decimal Notation

r = 6.82842 \dots, r = 1.17157 \dots

Solution steps

r^{2} - 8 r = - 8

Move

8

to the left side

r^{2} - 8 r + 8 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

(- 8)^{2} - 4 \cdot 1 \cdot 8 = 42

r_{1, 2} = \frac{- ( - 8 ) \pm 4 2}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 8 ) + 4 2}{2 \cdot 1}, r_{2} = \frac{- ( - 8 ) - 4 2}{2 \cdot 1}

r = \frac{- ( - 8 ) + 4 2}{2 \cdot 1} : 2 (2 + 2)

r = \frac{- ( - 8 ) - 4 2}{2 \cdot 1} : 2 (2 - 2)

The solutions to the quadratic equation are:

r = 2 (2 + 2), r = 2 (2 - 2)

12 x^{2} - 17 x + 6 = 0

- 3 x^{2} - 39 x + 133 = 9 x + 1

12 x^{2} - 184 x + 480 = 0

(5 x + 8)^{2} = 49

x^{2} - 6 x + 5 = 0

What are the solutions to the equation r^2-8r=-8 ?
The solutions to the equation r^2-8r=-8 are r=2(2+sqrt(2)),r=2(2-sqrt(2))
Find the zeros of r^2-8r=-8
The zeros of r^2-8r=-8 are r=2(2+sqrt(2)),r=2(2-sqrt(2))