.01e+.003e^2=7

Solution for .01e+.003e^2=7 equation:

.01e+.003e^2=7

We move all terms to the left:

.01e+.003e^2-(7)=0

a = .003; b = .01; c = -7;
Δ = b2-4ac
Δ = .012-4·.003·(-7)
Δ = 0.0841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$e_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.01)-\sqrt{0.0841}}{2*.003}=\frac{-0.01-\sqrt{0.0841}}{0.006}
$
$e_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.01)+\sqrt{0.0841}}{2*.003}=\frac{-0.01+\sqrt{0.0841}}{0.006}
$