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x\left(-5x-16\right)
Factor out x.
-5x^{2}-16x=0
Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.
x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}}}{2\left(-5\right)}
All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.
x=\frac{-\left(-16\right)±16}{2\left(-5\right)}
Take the square root of \left(-16\right)^{2}.
x=\frac{16±16}{2\left(-5\right)}
The opposite of -16 is 16.
x=\frac{16±16}{-10}
Multiply 2 times -5.
x=\frac{32}{-10}
Now solve the equation x=\frac{16±16}{-10} when ± is plus. Add 16 to 16.
x=-\frac{16}{5}
Reduce the fraction \frac{32}{-10} to lowest terms by extracting and canceling out 2.
x=\frac{0}{-10}
Now solve the equation x=\frac{16±16}{-10} when ± is minus. Subtract 16 from 16.
x=0
Divide 0 by -10.
-5x^{2}-16x=-5\left(x-\left(-\frac{16}{5}\right)\right)x
Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{16}{5} for x_{1} and 0 for x_{2}.
-5x^{2}-16x=-5\left(x+\frac{16}{5}\right)x
Simplify all the expressions of the form p-\left(-q\right) to p+q.
-5x^{2}-16x=-5\times \frac{-5x-16}{-5}x
Add \frac{16}{5} to x by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
-5x^{2}-16x=\left(-5x-16\right)x
Cancel out 5, the greatest common factor in -5 and -5.