What are the solutions to the equation x=-6x^2+8x+4 ?

The solutions to the equation x=-6x^2+8x+4 are x=-(-7+sqrt(145))/(12),x=(7+sqrt(145))/(12)

Find the zeros of x=-6x^2+8x+4

The zeros of x=-6x^2+8x+4 are x=-(-7+sqrt(145))/(12),x=(7+sqrt(145))/(12)

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x = - 6 x^{2} + 8 x + 4

x = - \frac{- 7 + 145}{12}, x = \frac{7 + 145}{12}

Decimal Notation

x = - 0.42013 \dots, x = 1.58679 \dots

Solution steps

x = - 6 x^{2} + 8 x + 4

Switch sides

- 6 x^{2} + 8 x + 4 = x

Move

x

to the left side

- 6 x^{2} + 7 x + 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 ( - 6 ) \cdot 4}{2 ( - 6 )}

7^{2} - 4 (- 6) \cdot 4 = 145

x_{1, 2} = \frac{- 7 \pm 145}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- 7 + 145}{2 ( - 6 )}, x_{2} = \frac{- 7 - 145}{2 ( - 6 )}

x = \frac{- 7 + 145}{2 ( - 6 )} : - \frac{- 7 + 145}{12}

x = \frac{- 7 - 145}{2 ( - 6 )} : \frac{7 + 145}{12}

The solutions to the quadratic equation are:

x = - \frac{- 7 + 145}{12}, x = \frac{7 + 145}{12}

x^{2} - 3 x + \frac{1}{4} = 0

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

x, x^{2} y + 3 x y^{3} - x = 3

9 x^{2} - 4 x = 1

9 u^{2} - u - 6534 = 0

What are the solutions to the equation x=-6x^2+8x+4 ?
The solutions to the equation x=-6x^2+8x+4 are x=-(-7+sqrt(145))/(12),x=(7+sqrt(145))/(12)
Find the zeros of x=-6x^2+8x+4
The zeros of x=-6x^2+8x+4 are x=-(-7+sqrt(145))/(12),x=(7+sqrt(145))/(12)