r^2+3r+0.07=0

Solution for r^2+3r+0.07=0 equation:

r^2+3r+0.07=0

a = 1; b = 3; c = +0.07;
Δ = b2-4ac
Δ = 32-4·1·0.07
Δ = 8.72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{8.72}}{2*1}=\frac{-3-\sqrt{8.72}}{2}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{8.72}}{2*1}=\frac{-3+\sqrt{8.72}}{2}
$