x^2+(-16x)=0

Solution for x^2+(-16x)=0 equation:

x^2+(-16x)=0

We get rid of parentheses

x^2-16x=0

a = 1; b = -16; c = 0;
Δ = b2-4ac
Δ = -162-4·1·0
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-16}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+16}{2*1}=\frac{32}{2}
=16
$