What are the solutions to the equation r^2+3r=6 ?

The solutions to the equation r^2+3r=6 are r=(-3+sqrt(33))/2 ,r=(-3-sqrt(33))/2

Find the zeros of r^2+3r=6

The zeros of r^2+3r=6 are r=(-3+sqrt(33))/2 ,r=(-3-sqrt(33))/2

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r^{2} + 3 r = 6

r = \frac{- 3 + 33}{2}, r = \frac{- 3 - 33}{2}

Decimal Notation

r = 1.37228 \dots, r = - 4.37228 \dots

Solution steps

r^{2} + 3 r = 6

Move

6

to the left side

r^{2} + 3 r - 6 = 0

Solve with the quadratic formula

r_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 6 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 6) = 33

r_{1, 2} = \frac{- 3 \pm 33}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 33}{2 \cdot 1}, r_{2} = \frac{- 3 - 33}{2 \cdot 1}

r = \frac{- 3 + 33}{2 \cdot 1} : \frac{- 3 + 33}{2}

r = \frac{- 3 - 33}{2 \cdot 1} : \frac{- 3 - 33}{2}

The solutions to the quadratic equation are:

r = \frac{- 3 + 33}{2}, r = \frac{- 3 - 33}{2}

\frac{3 x ^{2}}{2} + 4 x - \frac{2}{3} = 0

18 x^{2} - 16 x - 8 = - 16 x

x^{2} + 3 x - 26 = 0

d^{2} - 2 = 0

y, f = x^{4} y - 2 x^{3} y^{2}

What are the solutions to the equation r^2+3r=6 ?
The solutions to the equation r^2+3r=6 are r=(-3+sqrt(33))/2 ,r=(-3-sqrt(33))/2
Find the zeros of r^2+3r=6
The zeros of r^2+3r=6 are r=(-3+sqrt(33))/2 ,r=(-3-sqrt(33))/2