7t^2+28t+56=140

Solution for 7t^2+28t+56=140 equation:

7t^2+28t+56=140

We move all terms to the left:

7t^2+28t+56-(140)=0

We add all the numbers together, and all the variables

7t^2+28t-84=0

a = 7; b = 28; c = -84;
Δ = b2-4ac
Δ = 282-4·7·(-84)
Δ = 3136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3136}=56$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-56}{2*7}=\frac{-84}{14}
=-6
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+56}{2*7}=\frac{28}{14}
=2
$