What are the solutions to the equation quadraticformula 6-6x^2=3x ?

The solutions to the equation quadraticformula 6-6x^2=3x are x=-(1+sqrt(17))/4 ,x=(sqrt(17)-1)/4

Find the zeros of quadraticformula 6-6x^2=3x

The zeros of quadraticformula 6-6x^2=3x are x=-(1+sqrt(17))/4 ,x=(sqrt(17)-1)/4

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6 - 6 x^{2} = 3 x

x = - \frac{1 + 17}{4}, x = \frac{17 - 1}{4}

Decimal Notation

x = - 1.28077 \dots, x = 0.78077 \dots

Solution steps

6 - 6 x^{2} = 3 x

Move

3 x

to the left side

6 - 6 x^{2} - 3 x = 0

Write in the standard form

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

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 (- 6) \cdot 6 = 317

x_{1, 2} = \frac{- ( - 3 ) \pm 3 17}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 17}{2 ( - 6 )}, x_{2} = \frac{- ( - 3 ) - 3 17}{2 ( - 6 )}

x = \frac{- ( - 3 ) + 3 17}{2 ( - 6 )} : - \frac{1 + 17}{4}

x = \frac{- ( - 3 ) - 3 17}{2 ( - 6 )} : \frac{17 - 1}{4}

The solutions to the quadratic equation are:

x = - \frac{1 + 17}{4}, x = \frac{17 - 1}{4}

g, \frac{1}{6} g + 3 = 4 - \frac{1}{3} g

m, 4 - \frac{3}{5} m = \frac{6}{5} m + 2

(x - 2) (x + 5) = 18

x^{2} - 6 x = 40

- 37 - 45

What are the solutions to the equation quadraticformula 6-6x^2=3x ?
The solutions to the equation quadraticformula 6-6x^2=3x are x=-(1+sqrt(17))/4 ,x=(sqrt(17)-1)/4
Find the zeros of quadraticformula 6-6x^2=3x
The zeros of quadraticformula 6-6x^2=3x are x=-(1+sqrt(17))/4 ,x=(sqrt(17)-1)/4