(t+10)(t+20)+t=180

Solution for (t+10)(t+20)+t=180 equation:

(t+10)(t+20)+t=180

We move all terms to the left:

(t+10)(t+20)+t-(180)=0

We add all the numbers together, and all the variables

t+(t+10)(t+20)-180=0

We multiply parentheses ..

(+t^2+20t+10t+200)+t-180=0

We get rid of parentheses

t^2+20t+10t+t+200-180=0

We add all the numbers together, and all the variables

t^2+31t+20=0

a = 1; b = 31; c = +20;
Δ = b2-4ac
Δ = 312-4·1·20
Δ = 881
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{-(31)-\sqrt{881}}{2*1}=\frac{-31-\sqrt{881}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+\sqrt{881}}{2*1}=\frac{-31+\sqrt{881}}{2}
$