What are the solutions to the equation 3r^2+3r-4=0 ?

The solutions to the equation 3r^2+3r-4=0 are r=(-3+sqrt(57))/6 ,r=(-3-sqrt(57))/6

Find the zeros of 3r^2+3r-4=0

The zeros of 3r^2+3r-4=0 are r=(-3+sqrt(57))/6 ,r=(-3-sqrt(57))/6

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3 r^{2} + 3 r - 4 = 0

r = \frac{- 3 + 57}{6}, r = \frac{- 3 - 57}{6}

Solution steps

3 r^{2} + 3 r - 4 = 0

Solve with the quadratic formula

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

3^{2} - 4 \cdot 3 (- 4) = 57

r_{1, 2} = \frac{- 3 \pm 57}{2 \cdot 3}

Separate the solutions

r_{1} = \frac{- 3 + 57}{2 \cdot 3}, r_{2} = \frac{- 3 - 57}{2 \cdot 3}

r = \frac{- 3 + 57}{2 \cdot 3} : \frac{- 3 + 57}{6}

r = \frac{- 3 - 57}{2 \cdot 3} : \frac{- 3 - 57}{6}

The solutions to the quadratic equation are:

r = \frac{- 3 + 57}{6}, r = \frac{- 3 - 57}{6}

- 5 (2 x - 3)^{2} = - 125

x, x^{2} - 8 k x - 9 = 0

x, 5 x^{2} + x y^{2} = C

2 x^{2} + 3 x + 6

x^{2} - x - 20 = - x

What are the solutions to the equation 3r^2+3r-4=0 ?
The solutions to the equation 3r^2+3r-4=0 are r=(-3+sqrt(57))/6 ,r=(-3-sqrt(57))/6
Find the zeros of 3r^2+3r-4=0
The zeros of 3r^2+3r-4=0 are r=(-3+sqrt(57))/6 ,r=(-3-sqrt(57))/6