m^2+2.6m-3=0

Solution for m^2+2.6m-3=0 equation:

m^2+2.6m-3=0

a = 1; b = 2.6; c = -3;
Δ = b2-4ac
Δ = 2.62-4·1·(-3)
Δ = 18.76
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2.6)-\sqrt{18.76}}{2*1}=\frac{-2.6-\sqrt{18.76}}{2}
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.6)+\sqrt{18.76}}{2*1}=\frac{-2.6+\sqrt{18.76}}{2}
$