20m^2+19m+3=

Solution for 20m^2+19m+3= equation:

20m^2+19m+3=

We move all terms to the left:

20m^2+19m+3-()=0

We add all the numbers together, and all the variables

20m^2+19m=0

a = 20; b = 19; c = 0;
Δ = b2-4ac
Δ = 192-4·20·0
Δ = 361
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{361}=19$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-19}{2*20}=\frac{-38}{40}
=-19/20
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+19}{2*20}=\frac{0}{40}
=0
$