H(t)=(49+4.9t)(10-t)

Solution for H(t)=(49+4.9t)(10-t) equation:

(H)=(49+4.9H)(10-H)

We move all terms to the left:

(H)-((49+4.9H)(10-H))=0

We add all the numbers together, and all the variables

H-((4.9H+49)(-1H+10))=0

We multiply parentheses ..

-((-4H^2+40H-49H+490))+H=0


We calculate terms in parentheses: -((-4H^2+40H-49H+490)), so:

(-4H^2+40H-49H+490)

We get rid of parentheses

-4H^2+40H-49H+490

We add all the numbers together, and all the variables

-4H^2-9H+490

Back to the equation:

-(-4H^2-9H+490)


We get rid of parentheses

4H^2+9H+H-490=0

We add all the numbers together, and all the variables

4H^2+10H-490=0

a = 4; b = 10; c = -490;
Δ = b2-4ac
Δ = 102-4·4·(-490)
Δ = 7940
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{7940}=\sqrt{4*1985}=\sqrt{4}*\sqrt{1985}=2\sqrt{1985}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{1985}}{2*4}=\frac{-10-2\sqrt{1985}}{8}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{1985}}{2*4}=\frac{-10+2\sqrt{1985}}{8}
$