(6t-2t^2)+18(7t+2)=7t

Solution for (6t-2t^2)+18(7t+2)=7t equation:

(6t-2t^2)+18(7t+2)=7t

We move all terms to the left:

(6t-2t^2)+18(7t+2)-(7t)=0

We add all the numbers together, and all the variables

(6t-2t^2)-7t+18(7t+2)=0

We multiply parentheses

(6t-2t^2)-7t+126t+36=0

We get rid of parentheses

-2t^2+6t-7t+126t+36=0

We add all the numbers together, and all the variables

-2t^2+125t+36=0

a = -2; b = 125; c = +36;
Δ = b2-4ac
Δ = 1252-4·(-2)·36
Δ = 15913
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{-(125)-\sqrt{15913}}{2*-2}=\frac{-125-\sqrt{15913}}{-4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(125)+\sqrt{15913}}{2*-2}=\frac{-125+\sqrt{15913}}{-4}
$