What are the solutions to the equation r^2+12r+4=7 ?

The solutions to the equation r^2+12r+4=7 are r=-6+sqrt(39),r=-6-sqrt(39)

Find the zeros of r^2+12r+4=7

The zeros of r^2+12r+4=7 are r=-6+sqrt(39),r=-6-sqrt(39)

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r^{2} + 12 r + 4 = 7

r = - 6 + 39, r = - 6 - 39

Decimal Notation

r = 0.24499 \dots, r = - 12.24499 \dots

Solution steps

r^{2} + 12 r + 4 = 7

Move

7

to the left side

r^{2} + 12 r - 3 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 3 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 3) = 239

r_{1, 2} = \frac{- 12 \pm 2 39}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 12 + 2 39}{2 \cdot 1}, r_{2} = \frac{- 12 - 2 39}{2 \cdot 1}

r = \frac{- 12 + 2 39}{2 \cdot 1} : - 6 + 39

r = \frac{- 12 - 2 39}{2 \cdot 1} : - 6 - 39

The solutions to the quadratic equation are:

r = - 6 + 39, r = - 6 - 39

- 5 x^{2} - 25 = 0

2 x^{2} - 4 x + 1 = 2 x - 3 - x^{2}

9 = - a^{2} + 10 a

x^{2} - 4 a^{2} - b^{2} = 4 ab

m x^{2} - 2 (m + 1) x + 2 m = 0

What are the solutions to the equation r^2+12r+4=7 ?
The solutions to the equation r^2+12r+4=7 are r=-6+sqrt(39),r=-6-sqrt(39)
Find the zeros of r^2+12r+4=7
The zeros of r^2+12r+4=7 are r=-6+sqrt(39),r=-6-sqrt(39)