-5=(t+15)(t+20)

Solution for -5=(t+15)(t+20) equation:

-5=(t+15)(t+20)

We move all terms to the left:

-5-((t+15)(t+20))=0

We multiply parentheses ..

-((+t^2+20t+15t+300))-5=0


We calculate terms in parentheses: -((+t^2+20t+15t+300)), so:

(+t^2+20t+15t+300)

We get rid of parentheses

t^2+20t+15t+300

We add all the numbers together, and all the variables

t^2+35t+300

Back to the equation:

-(t^2+35t+300)


We get rid of parentheses

-t^2-35t-300-5=0

We add all the numbers together, and all the variables

-1t^2-35t-305=0

a = -1; b = -35; c = -305;
Δ = b2-4ac
Δ = -352-4·(-1)·(-305)
Δ = 5
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{-(-35)-\sqrt{5}}{2*-1}=\frac{35-\sqrt{5}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+\sqrt{5}}{2*-1}=\frac{35+\sqrt{5}}{-2}
$