What are the solutions to the equation -5-18x+6x^2=0 ?

The solutions to the equation -5-18x+6x^2=0 are x=(9+sqrt(111))/6 ,x=(9-sqrt(111))/6

Find the zeros of -5-18x+6x^2=0

The zeros of -5-18x+6x^2=0 are x=(9+sqrt(111))/6 ,x=(9-sqrt(111))/6

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- 5 - 18 x + 6 x^{2} = 0

x = \frac{9 + 111}{6}, x = \frac{9 - 111}{6}

Decimal Notation

x = 3.25594 \dots, x = - 0.25594 \dots

Solution steps

- 5 - 18 x + 6 x^{2} = 0

Write in the standard form

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

6 x^{2} - 18 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

(- 18)^{2} - 4 \cdot 6 (- 5) = 2111

x_{1, 2} = \frac{- ( - 18 ) \pm 2 111}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 2 111}{2 \cdot 6}, x_{2} = \frac{- ( - 18 ) - 2 111}{2 \cdot 6}

x = \frac{- ( - 18 ) + 2 111}{2 \cdot 6} : \frac{9 + 111}{6}

x = \frac{- ( - 18 ) - 2 111}{2 \cdot 6} : \frac{9 - 111}{6}

The solutions to the quadratic equation are:

x = \frac{9 + 111}{6}, x = \frac{9 - 111}{6}

x^{2} + 3 x - 14 = 0

∣ x + 2 ∣ \leq 6

(- 8 + 2 i) (- 8 + 5 i)

\frac{10}{1 0 ^{\frac{1}{4}}}

∣ 2 x - 3 ∣ \leq 11

What are the solutions to the equation -5-18x+6x^2=0 ?
The solutions to the equation -5-18x+6x^2=0 are x=(9+sqrt(111))/6 ,x=(9-sqrt(111))/6
Find the zeros of -5-18x+6x^2=0
The zeros of -5-18x+6x^2=0 are x=(9+sqrt(111))/6 ,x=(9-sqrt(111))/6