What are the solutions to the equation 36-3x+4-x^2=47-2x^2+9x ?

The solutions to the equation 36-3x+4-x^2=47-2x^2+9x are x=6+sqrt(43),x=6-sqrt(43)

Find the zeros of 36-3x+4-x^2=47-2x^2+9x

The zeros of 36-3x+4-x^2=47-2x^2+9x are x=6+sqrt(43),x=6-sqrt(43)

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36 - 3 x + 4 - x^{2} = 47 - 2 x^{2} + 9 x

x = 6 + 43, x = 6 - 43

Decimal Notation

x = 12.55743 \dots, x = - 0.55743 \dots

Solution steps

36 - 3 x + 4 - x^{2} = 47 - 2 x^{2} + 9 x

Refine

- x^{2} - 3 x + 40 = 47 - 2 x^{2} + 9 x

Move

9 x

to the left side

- x^{2} - 12 x + 40 = 47 - 2 x^{2}

Move

2 x^{2}

to the left side

x^{2} - 12 x + 40 = 47

Move

47

to the left side

x^{2} - 12 x - 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 1 \cdot ( - 7 )}{2 \cdot 1}

(- 12)^{2} - 4 \cdot 1 \cdot (- 7) = 243

x_{1, 2} = \frac{- ( - 12 ) \pm 2 43}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 2 43}{2 \cdot 1}, x_{2} = \frac{- ( - 12 ) - 2 43}{2 \cdot 1}

x = \frac{- ( - 12 ) + 2 43}{2 \cdot 1} : 6 + 43

x = \frac{- ( - 12 ) - 2 43}{2 \cdot 1} : 6 - 43

The solutions to the quadratic equation are:

x = 6 + 43, x = 6 - 43

\frac{x + 4}{- 2} \leq \frac{3 x - 8}{3}

2 x^{2} - 19 x - 33 = 0

(2 x + 3) (2 x - 3) + 5 x = 2 (x + 1) - 1

x^{4} - x^{3} - 13 x^{2} + x + 12

3 x^{2} + 25 x + 8

What are the solutions to the equation 36-3x+4-x^2=47-2x^2+9x ?
The solutions to the equation 36-3x+4-x^2=47-2x^2+9x are x=6+sqrt(43),x=6-sqrt(43)
Find the zeros of 36-3x+4-x^2=47-2x^2+9x
The zeros of 36-3x+4-x^2=47-2x^2+9x are x=6+sqrt(43),x=6-sqrt(43)