Question

sole x^2 +20x=-20

Answer 1

To solve the equation \( x^2 + 20x = -20 \), we first rearrange the equation to set it to zero:

\[
x^2 + 20x + 20 = 0
\]

Now, we can use the quadratic formula to solve for \( x \). The quadratic formula is given by:

\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\]

In our equation, \( a = 1 \), \( b = 20 \), and \( c = 20 \).

Next, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 20^2 - 4 \cdot 1 \cdot 20 = 400 - 80 = 320
\]

Now we can find the roots using the quadratic formula:

\[
x = \frac{{-20 \pm \sqrt{320}}}{2 \cdot 1}
\]

First, simplify \( \sqrt{320} \):

\[
\sqrt{320} = \sqrt{64 \cdot 5} = 8\sqrt{5}
\]

Now substitute this back into the formula:

\[
x = \frac{{-20 \pm 8\sqrt{5}}}{2}
\]

This simplifies to:

\[
x = -10 \pm 4\sqrt{5}
\]

Thus, the solutions for the equation \( x^2 + 20x + 20 = 0 \) are:

\[
x_1 = -10 + 4\sqrt{5}
\]
\[
x_2 = -10 - 4\sqrt{5}
\]