What are the solutions to the equation x^2+3x-68=3x-43x-4 ?

The solutions to the equation x^2+3x-68=3x-43x-4 are x=(-43+sqrt(2105))/2 ,x=(-43-sqrt(2105))/2

Find the zeros of x^2+3x-68=3x-43x-4

The zeros of x^2+3x-68=3x-43x-4 are x=(-43+sqrt(2105))/2 ,x=(-43-sqrt(2105))/2

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x^{2} + 3 x - 68 = 3 x - 43 x - 4

x = \frac{- 43 + 2105}{2}, x = \frac{- 43 - 2105}{2}

Decimal Notation

x = 1.44013 \dots, x = - 44.44013 \dots

Solution steps

x^{2} + 3 x - 68 = 3 x - 43 x - 4

Refine

x^{2} + 3 x - 68 = - 40 x - 4

Move

4

to the left side

x^{2} + 3 x - 64 = - 40 x

Move

40 x

to the left side

x^{2} + 43 x - 64 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 43 \pm 4 3 ^{2} - 4 \cdot 1 \cdot ( - 64 )}{2 \cdot 1}

4 3^{2} - 4 \cdot 1 \cdot (- 64) = 2105

x_{1, 2} = \frac{- 43 \pm 2105}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 43 + 2105}{2 \cdot 1}, x_{2} = \frac{- 43 - 2105}{2 \cdot 1}

x = \frac{- 43 + 2105}{2 \cdot 1} : \frac{- 43 + 2105}{2}

x = \frac{- 43 - 2105}{2 \cdot 1} : \frac{- 43 - 2105}{2}

The solutions to the quadratic equation are:

x = \frac{- 43 + 2105}{2}, x = \frac{- 43 - 2105}{2}

9 w^{2} + 4 = 0

9 x - 4 x^{2} = 7

x^{2} - 18 x + 3 = 0

2 (x - 2)^{2} = 288

2 x (x + 1) = 2

What are the solutions to the equation x^2+3x-68=3x-43x-4 ?
The solutions to the equation x^2+3x-68=3x-43x-4 are x=(-43+sqrt(2105))/2 ,x=(-43-sqrt(2105))/2
Find the zeros of x^2+3x-68=3x-43x-4
The zeros of x^2+3x-68=3x-43x-4 are x=(-43+sqrt(2105))/2 ,x=(-43-sqrt(2105))/2