H(s)=0.003s^2+0.07s-0.027

Solution for H(s)=0.003s^2+0.07s-0.027 equation:

(H)=0.003H^2+0.07H-0.027

We move all terms to the left:

(H)-(0.003H^2+0.07H-0.027)=0

We get rid of parentheses

-0.003H^2+H-0.07H+0.027=0

We add all the numbers together, and all the variables

-0.003H^2+0.93H+0.027=0

a = -0.003; b = 0.93; c = +0.027;
Δ = b2-4ac
Δ = 0.932-4·(-0.003)·0.027
Δ = 0.865224
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.93)-\sqrt{0.865224}}{2*-0.003}=\frac{-0.93-\sqrt{0.865224}}{-0.006}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.93)+\sqrt{0.865224}}{2*-0.003}=\frac{-0.93+\sqrt{0.865224}}{-0.006}
$