H(3.1)=16t^2+160t+22

Solution for H(3.1)=16t^2+160t+22 equation:

(3.1)=16H^2+160H+22

We move all terms to the left:

(3.1)-(16H^2+160H+22)=0

We add all the numbers together, and all the variables

-(16H^2+160H+22)+3.1=0

We get rid of parentheses

-16H^2-160H-22+3.1=0

We add all the numbers together, and all the variables

-16H^2-160H-18.9=0

a = -16; b = -160; c = -18.9;
Δ = b2-4ac
Δ = -1602-4·(-16)·(-18.9)
Δ = 24390.4
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{-(-160)-\sqrt{24390.4}}{2*-16}=\frac{160-\sqrt{24390.4}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+\sqrt{24390.4}}{2*-16}=\frac{160+\sqrt{24390.4}}{-32}
$