What are the solutions to the equation 180=x(3+x) ?

The solutions to the equation 180=x(3+x) are x=12,x=-15

Find the zeros of 180=x(3+x)

The zeros of 180=x(3+x) are x=12,x=-15

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180 = x (3 + x)

x = 12, x = - 15

Solution steps

180 = x (3 + x)

Expand

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

180 = 3 x + x^{2}

Switch sides

3 x + x^{2} = 180

Move

180

to the left side

3 x + x^{2} - 180 = 0

Write in the standard form

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

x^{2} + 3 x - 180 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 1 \cdot ( - 180 )}{2 \cdot 1}

3^{2} - 4 \cdot 1 \cdot (- 180) = 27

x_{1, 2} = \frac{- 3 \pm 27}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 3 + 27}{2 \cdot 1}, x_{2} = \frac{- 3 - 27}{2 \cdot 1}

x = \frac{- 3 + 27}{2 \cdot 1} : 12

x = \frac{- 3 - 27}{2 \cdot 1} : - 15

The solutions to the quadratic equation are:

x = 12, x = - 15

x, f^{- 1} (x) = x^{2} + 4

x, f^{- 1} (x) = x^{2} - 1

2 x [- 2 (- x + 2) + 4] = (x + 2) (x - 2)

x^{2} - 14 x + 21 = 0

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

What are the solutions to the equation 180=x(3+x) ?
The solutions to the equation 180=x(3+x) are x=12,x=-15
Find the zeros of 180=x(3+x)
The zeros of 180=x(3+x) are x=12,x=-15