23m^2+100m+19=0

Solution for 23m^2+100m+19=0 equation:

23m^2+100m+19=0

a = 23; b = 100; c = +19;
Δ = b2-4ac
Δ = 1002-4·23·19
Δ = 8252
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{8252}=\sqrt{4*2063}=\sqrt{4}*\sqrt{2063}=2\sqrt{2063}$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-2\sqrt{2063}}{2*23}=\frac{-100-2\sqrt{2063}}{46}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+2\sqrt{2063}}{2*23}=\frac{-100+2\sqrt{2063}}{46}
$