10x^2+20x+x+2=0

Solution for 10x^2+20x+x+2=0 equation:

10x^2+20x+x+2=0

We add all the numbers together, and all the variables

10x^2+21x+2=0

a = 10; b = 21; c = +2;
Δ = b2-4ac
Δ = 212-4·10·2
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{-(21)-19}{2*10}=\frac{-40}{20}
=-2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+19}{2*10}=\frac{-2}{20}
=-1/10
$