(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+10t-5=0

a = 21; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·21·(-5)
Δ = 520
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:

$\sqrt{\Delta}=\sqrt{520}=\sqrt{4*130}=\sqrt{4}*\sqrt{130}=2\sqrt{130}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{130}}{2*21}=\frac{-10-2\sqrt{130}}{42}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{130}}{2*21}=\frac{-10+2\sqrt{130}}{42}
$