What are the solutions to the equation solvefor x,600-5-7=3x(x+10) ?

The solutions to the equation solvefor x,600-5-7=3x(x+10) are x=-5+sqrt(221),x=-5-sqrt(221)

Find the zeros of solvefor x,600-5-7=3x(x+10)

The zeros of solvefor x,600-5-7=3x(x+10) are x=-5+sqrt(221),x=-5-sqrt(221)

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x, 600 - 5 - 7 = 3 x (x + 10)

x = - 5 + 221, x = - 5 - 221

Decimal Notation

x = 9.86606 \dots, x = - 19.86606 \dots

Solution steps

600 - 5 - 7 = 3 x (x + 10)

Expand

3 x (x + 10) : 3 x^{2} + 30 x

588 = 3 x^{2} + 30 x

Switch sides

3 x^{2} + 30 x = 588

Move

588

to the left side

3 x^{2} + 30 x - 588 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 30 \pm 3 0 ^{2} - 4 \cdot 3 ( - 588 )}{2 \cdot 3}

3 0^{2} - 4 \cdot 3 (- 588) = 6221

x_{1, 2} = \frac{- 30 \pm 6 221}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 30 + 6 221}{2 \cdot 3}, x_{2} = \frac{- 30 - 6 221}{2 \cdot 3}

x = \frac{- 30 + 6 221}{2 \cdot 3} : - 5 + 221

x = \frac{- 30 - 6 221}{2 \cdot 3} : - 5 - 221

The solutions to the quadratic equation are:

x = - 5 + 221, x = - 5 - 221

x^{2} + 10 x = - 4 x - 40

x^{2} + x

x^{2} + 8

2 x^{2} + 5 x - 6 = - 3

5 r^{2} - 12 r + 6 = 2

What are the solutions to the equation solvefor x,600-5-7=3x(x+10) ?
The solutions to the equation solvefor x,600-5-7=3x(x+10) are x=-5+sqrt(221),x=-5-sqrt(221)
Find the zeros of solvefor x,600-5-7=3x(x+10)
The zeros of solvefor x,600-5-7=3x(x+10) are x=-5+sqrt(221),x=-5-sqrt(221)