(t+2)(14+3t)=0

Solution for (t+2)(14+3t)=0 equation:

(t+2)(14+3t)=0

We add all the numbers together, and all the variables

(t+2)(3t+14)=0

We multiply parentheses ..

(+3t^2+14t+6t+28)=0

We get rid of parentheses

3t^2+14t+6t+28=0

We add all the numbers together, and all the variables

3t^2+20t+28=0

a = 3; b = 20; c = +28;
Δ = b2-4ac
Δ = 202-4·3·28
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-8}{2*3}=\frac{-28}{6}
=-4+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+8}{2*3}=\frac{-12}{6}
=-2
$