What are the solutions to the equation r^2+14r+20=19 ?

The solutions to the equation r^2+14r+20=19 are r=-7+4sqrt(3),r=-7-4sqrt(3)

Find the zeros of r^2+14r+20=19

The zeros of r^2+14r+20=19 are r=-7+4sqrt(3),r=-7-4sqrt(3)

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r^{2} + 14 r + 20 = 19

r = - 7 + 43, r = - 7 - 43

Decimal Notation

r = - 0.07179 \dots, r = - 13.92820 \dots

Solution steps

r^{2} + 14 r + 20 = 19

Move

19

to the left side

r^{2} + 14 r + 1 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 14 \pm 1 4 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

1 4^{2} - 4 \cdot 1 \cdot 1 = 83

r_{1, 2} = \frac{- 14 \pm 8 3}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 14 + 8 3}{2 \cdot 1}, r_{2} = \frac{- 14 - 8 3}{2 \cdot 1}

r = \frac{- 14 + 8 3}{2 \cdot 1} : - 7 + 43

r = \frac{- 14 - 8 3}{2 \cdot 1} : - 7 - 43

The solutions to the quadratic equation are:

r = - 7 + 43, r = - 7 - 43

\frac{1}{3} x + \frac{2}{3} (x - 24) = \frac{1}{6} x - 6

6 x^{2} - 150 = 0

2^{4} \cdot 2^{2}

\frac{7 x}{3} + \frac{13}{2} - \frac{7 x}{6} = \frac{17}{12} - \frac{3 x}{4}

- 2 = x + 14

What are the solutions to the equation r^2+14r+20=19 ?
The solutions to the equation r^2+14r+20=19 are r=-7+4sqrt(3),r=-7-4sqrt(3)
Find the zeros of r^2+14r+20=19
The zeros of r^2+14r+20=19 are r=-7+4sqrt(3),r=-7-4sqrt(3)