16x^2+18x+x=0

Solution for 16x^2+18x+x=0 equation:

16x^2+18x+x=0

We add all the numbers together, and all the variables

16x^2+19x=0

a = 16; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·16·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*16}=\frac{-38}{32}
=-1+3/16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*16}=\frac{0}{32}
=0
$