(T+20)=(t+10)t

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

(+20)=(T+10)T

We move all terms to the left:

(+20)-((T+10)T)=0

We add all the numbers together, and all the variables

-((T+10)T)+20=0


We calculate terms in parentheses: -((T+10)T), so:

(T+10)T

We multiply parentheses

T^2+10T

Back to the equation:

-(T^2+10T)


We get rid of parentheses

-T^2-10T+20=0

We add all the numbers together, and all the variables

-1T^2-10T+20=0

a = -1; b = -10; c = +20;
Δ = b2-4ac
Δ = -102-4·(-1)·20
Δ = 180
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$T_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-6\sqrt{5}}{2*-1}=\frac{10-6\sqrt{5}}{-2}
$
$T_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+6\sqrt{5}}{2*-1}=\frac{10+6\sqrt{5}}{-2}
$