(7t-2)(3t+-1)=3(1-3t)

Solution for (7t-2)(3t+-1)=3(1-3t) equation:

(7t-2)(3t+-1)=3(1-3t)

We move all terms to the left:

(7t-2)(3t+-1)-(3(1-3t))=0

We add all the numbers together, and all the variables

(7t-2)(3t-1)-(3(-3t+1))=0

We multiply parentheses ..

(+21t^2-7t-6t+2)-(3(-3t+1))=0


We calculate terms in parentheses: -(3(-3t+1)), so:

3(-3t+1)

We multiply parentheses

-9t+3

Back to the equation:

-(-9t+3)


We get rid of parentheses

21t^2-7t-6t+9t+2-3=0

We add all the numbers together, and all the variables

21t^2-4t-1=0

a = 21; b = -4; c = -1;
Δ = b2-4ac
Δ = -42-4·21·(-1)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-10}{2*21}=\frac{-6}{42}
=-1/7
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+10}{2*21}=\frac{14}{42}
=1/3
$