(7t+1)(t-2)=0

Solution for (7t+1)(t-2)=0 equation:

(7t+1)(t-2)=0

We multiply parentheses ..

(+7t^2-14t+t-2)=0

We get rid of parentheses

7t^2-14t+t-2=0

We add all the numbers together, and all the variables

7t^2-13t-2=0

a = 7; b = -13; c = -2;
Δ = b2-4ac
Δ = -132-4·7·(-2)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-15}{2*7}=\frac{-2}{14}
=-1/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+15}{2*7}=\frac{28}{14}
=2
$