.0003x^2+.1x-43=0

Solution for .0003x^2+.1x-43=0 equation:

.0003x^2+.1x-43=0

a = .0003; b = .1; c = -43;
Δ = b2-4ac
Δ = .12-4·.0003·(-43)
Δ = 0.0616
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(.1)-\sqrt{0.0616}}{2*.0003}=\frac{-0.1-\sqrt{0.0616}}{0.0006}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(.1)+\sqrt{0.0616}}{2*.0003}=\frac{-0.1+\sqrt{0.0616}}{0.0006}
$