What are the solutions to the equation x(x+12)=400 ?

The solutions to the equation x(x+12)=400 are x=2(sqrt(109)-3),x=-2(3+sqrt(109))

Find the zeros of x(x+12)=400

The zeros of x(x+12)=400 are x=2(sqrt(109)-3),x=-2(3+sqrt(109))

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x (x + 12) = 400

x = 2 (109 - 3), x = - 2 (3 + 109)

Decimal Notation

x = 14.88061 \dots, x = - 26.88061 \dots

Solution steps

x (x + 12) = 400

Expand

x (x + 12) : x^{2} + 12 x

x^{2} + 12 x = 400

Move

400

to the left side

x^{2} + 12 x - 400 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 400 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 400) = 4109

x_{1, 2} = \frac{- 12 \pm 4 109}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 12 + 4 109}{2 \cdot 1}, x_{2} = \frac{- 12 - 4 109}{2 \cdot 1}

x = \frac{- 12 + 4 109}{2 \cdot 1} : 2 (109 - 3)

x = \frac{- 12 - 4 109}{2 \cdot 1} : - 2 (3 + 109)

The solutions to the quadratic equation are:

x = 2 (109 - 3), x = - 2 (3 + 109)

x, x^{2} + 4 x y + y^{2} = 9

2 X^{2} - 6 X = 6 X^{2} - 8 X

4 p^{2} - 12 p + 7 = 0

4 e^{2} + 3 e = - 12

y, (x + y)^{3} = x^{3} + y^{3}

What are the solutions to the equation x(x+12)=400 ?
The solutions to the equation x(x+12)=400 are x=2(sqrt(109)-3),x=-2(3+sqrt(109))
Find the zeros of x(x+12)=400
The zeros of x(x+12)=400 are x=2(sqrt(109)-3),x=-2(3+sqrt(109))