What are the solutions to the equation r^2-16r+47=-5 ?

The solutions to the equation r^2-16r+47=-5 are r=2(4+sqrt(3)),r=2(4-sqrt(3))

Find the zeros of r^2-16r+47=-5

The zeros of r^2-16r+47=-5 are r=2(4+sqrt(3)),r=2(4-sqrt(3))

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r^{2} - 16 r + 47 = - 5

r = 2 (4 + 3), r = 2 (4 - 3)

Decimal Notation

r = 11.46410 \dots, r = 4.53589 \dots

Solution steps

r^{2} - 16 r + 47 = - 5

Move

5

to the left side

r^{2} - 16 r + 52 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot 52}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot 52 = 43

r_{1, 2} = \frac{- ( - 16 ) \pm 4 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- ( - 16 ) + 4 3}{2 \cdot 1}, r_{2} = \frac{- ( - 16 ) - 4 3}{2 \cdot 1}

r = \frac{- ( - 16 ) + 4 3}{2 \cdot 1} : 2 (4 + 3)

r = \frac{- ( - 16 ) - 4 3}{2 \cdot 1} : 2 (4 - 3)

The solutions to the quadratic equation are:

r = 2 (4 + 3), r = 2 (4 - 3)

x - 2 < 15

4 \times 2 - 1

\frac{1}{3} x < - 3 or 2 x > 0

3 x + 1 = - x + 5

x^{2} + 13 x + 58 = - 2 x + 8

What are the solutions to the equation r^2-16r+47=-5 ?
The solutions to the equation r^2-16r+47=-5 are r=2(4+sqrt(3)),r=2(4-sqrt(3))
Find the zeros of r^2-16r+47=-5
The zeros of r^2-16r+47=-5 are r=2(4+sqrt(3)),r=2(4-sqrt(3))