(3t+7)(t-2)=5

Solution for (3t+7)(t-2)=5 equation:

(3t+7)(t-2)=5

We move all terms to the left:

(3t+7)(t-2)-(5)=0

We multiply parentheses ..

(+3t^2-6t+7t-14)-5=0

We get rid of parentheses

3t^2-6t+7t-14-5=0

We add all the numbers together, and all the variables

3t^2+t-19=0

a = 3; b = 1; c = -19;
Δ = b2-4ac
Δ = 12-4·3·(-19)
Δ = 229
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{-(1)-\sqrt{229}}{2*3}=\frac{-1-\sqrt{229}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{229}}{2*3}=\frac{-1+\sqrt{229}}{6}
$