What are the solutions to the equation -20-3x+3x^2=0 ?

The solutions to the equation -20-3x+3x^2=0 are x=(3+sqrt(249))/6 ,x=(3-sqrt(249))/6

Find the zeros of -20-3x+3x^2=0

The zeros of -20-3x+3x^2=0 are x=(3+sqrt(249))/6 ,x=(3-sqrt(249))/6

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- 20 - 3 x + 3 x^{2} = 0

x = \frac{3 + 249}{6}, x = \frac{3 - 249}{6}

Solution steps

- 20 - 3 x + 3 x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} - 3 x - 20 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 3 ( - 20 )}{2 \cdot 3}

(- 3)^{2} - 4 \cdot 3 (- 20) = 249

x_{1, 2} = \frac{- ( - 3 ) \pm 249}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 249}{2 \cdot 3}, x_{2} = \frac{- ( - 3 ) - 249}{2 \cdot 3}

x = \frac{- ( - 3 ) + 249}{2 \cdot 3} : \frac{3 + 249}{6}

x = \frac{- ( - 3 ) - 249}{2 \cdot 3} : \frac{3 - 249}{6}

The solutions to the quadratic equation are:

x = \frac{3 + 249}{6}, x = \frac{3 - 249}{6}

(2 x - 5)^{2} - 81 = 0

4 x^{2} + 23 x - 72 = 0

x + 30 = 4 x^{2} + 20 x + 25

16 - 24 x + 9 x^{2} = 0

3 q^{2} - 136 q + 1300 = 0

What are the solutions to the equation -20-3x+3x^2=0 ?
The solutions to the equation -20-3x+3x^2=0 are x=(3+sqrt(249))/6 ,x=(3-sqrt(249))/6
Find the zeros of -20-3x+3x^2=0
The zeros of -20-3x+3x^2=0 are x=(3+sqrt(249))/6 ,x=(3-sqrt(249))/6