120=t(140+t)

Solution for 120=t(140+t) equation:

120=t(140+t)

We move all terms to the left:

120-(t(140+t))=0

We add all the numbers together, and all the variables

-(t(t+140))+120=0


We calculate terms in parentheses: -(t(t+140)), so:

t(t+140)

We multiply parentheses

t^2+140t

Back to the equation:

-(t^2+140t)


We get rid of parentheses

-t^2-140t+120=0

We add all the numbers together, and all the variables

-1t^2-140t+120=0

a = -1; b = -140; c = +120;
Δ = b2-4ac
Δ = -1402-4·(-1)·120
Δ = 20080
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:

$\sqrt{\Delta}=\sqrt{20080}=\sqrt{16*1255}=\sqrt{16}*\sqrt{1255}=4\sqrt{1255}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-140)-4\sqrt{1255}}{2*-1}=\frac{140-4\sqrt{1255}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-140)+4\sqrt{1255}}{2*-1}=\frac{140+4\sqrt{1255}}{-2}
$