-0.03x^2+2.3x-20=0

Solution for -0.03x^2+2.3x-20=0 equation:

-0.03x^2+2.3x-20=0

a = -0.03; b = 2.3; c = -20;
Δ = b2-4ac
Δ = 2.32-4·(-0.03)·(-20)
Δ = 2.89
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{-(2.3)-\sqrt{2.89}}{2*-0.03}=\frac{-2.3-\sqrt{2.89}}{-0.06}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2.3)+\sqrt{2.89}}{2*-0.03}=\frac{-2.3+\sqrt{2.89}}{-0.06}
$