(3t+4)(t+3)=7

Solution for (3t+4)(t+3)=7 equation:

(3t+4)(t+3)=7

We move all terms to the left:

(3t+4)(t+3)-(7)=0

We multiply parentheses ..

(+3t^2+9t+4t+12)-7=0

We get rid of parentheses

3t^2+9t+4t+12-7=0

We add all the numbers together, and all the variables

3t^2+13t+5=0

a = 3; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·3·5
Δ = 109
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{-(13)-\sqrt{109}}{2*3}=\frac{-13-\sqrt{109}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+\sqrt{109}}{2*3}=\frac{-13+\sqrt{109}}{6}
$