What are the solutions to the equation r^2+3r-3=0 ?

The solutions to the equation r^2+3r-3=0 are r=(-3+sqrt(21))/2 ,r=(-3-sqrt(21))/2

Find the zeros of r^2+3r-3=0

The zeros of r^2+3r-3=0 are r=(-3+sqrt(21))/2 ,r=(-3-sqrt(21))/2

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

r = \frac{- 3 + 21}{2}, r = \frac{- 3 - 21}{2}

Decimal Notation

r = 0.79128 \dots, r = - 3.79128 \dots

Solution steps

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

Solve with the quadratic formula

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

3^{2} - 4 \cdot 1 \cdot (- 3) = 21

r_{1, 2} = \frac{- 3 \pm 21}{2 \cdot 1}

Separate the solutions

r_{1} = \frac{- 3 + 21}{2 \cdot 1}, r_{2} = \frac{- 3 - 21}{2 \cdot 1}

r = \frac{- 3 + 21}{2 \cdot 1} : \frac{- 3 + 21}{2}

r = \frac{- 3 - 21}{2 \cdot 1} : \frac{- 3 - 21}{2}

The solutions to the quadratic equation are:

r = \frac{- 3 + 21}{2}, r = \frac{- 3 - 21}{2}

x^{4} + 2 x^{2} y^{2} + y^{4}

3 x + 1 = 0

(\frac{13}{3})^{2}

14 x^{2} y^{5} (11 x^{5} y^{5})

- \frac{1}{8} + \frac{3}{4}

What are the solutions to the equation r^2+3r-3=0 ?
The solutions to the equation r^2+3r-3=0 are r=(-3+sqrt(21))/2 ,r=(-3-sqrt(21))/2
Find the zeros of r^2+3r-3=0
The zeros of r^2+3r-3=0 are r=(-3+sqrt(21))/2 ,r=(-3-sqrt(21))/2