.96r^2+1.9104r+.9504=0

Solution for .96r^2+1.9104r+.9504=0 equation:

.96r^2+1.9104r+.9504=0

We add all the numbers together, and all the variables

.96r^2+1.9104r+0.9504=0

a = .96; b = 1.9104; c = +0.9504;
Δ = b2-4ac
Δ = 1.91042-4·.96·0.9504
Δ = 9.21600000003E-5
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{-(1.9104)-\sqrt{9.21600000003E-5}}{2*.96}=\frac{-1.9104-\sqrt{9.21600000003E-5}}{1.92}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1.9104)+\sqrt{9.21600000003E-5}}{2*.96}=\frac{-1.9104+\sqrt{9.21600000003E-5}}{1.92}
$