0=(t+0.2)(t+5.9)

Solution for 0=(t+0.2)(t+5.9) equation:

0=(t+0.2)(t+5.9)

We move all terms to the left:

0-((t+0.2)(t+5.9))=0

We add all the numbers together, and all the variables

-((t+0.2)(t+5.9))=0

We multiply parentheses ..

-((+t^2+5.9t+0.2t+1.18))=0


We calculate terms in parentheses: -((+t^2+5.9t+0.2t+1.18)), so:

(+t^2+5.9t+0.2t+1.18)

We get rid of parentheses

t^2+5.9t+0.2t+1.18

We add all the numbers together, and all the variables

t^2+6.1t+1.18

Back to the equation:

-(t^2+6.1t+1.18)


We get rid of parentheses

-t^2-6.1t-1.18=0

We add all the numbers together, and all the variables

-1t^2-6.1t-1.18=0

a = -1; b = -6.1; c = -1.18;
Δ = b2-4ac
Δ = -6.12-4·(-1)·(-1.18)
Δ = 32.49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6.1)-\sqrt{32.49}}{2*-1}=\frac{6.1-\sqrt{32.49}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6.1)+\sqrt{32.49}}{2*-1}=\frac{6.1+\sqrt{32.49}}{-2}
$