(t-2)(3t-10)=0

Solution for (t-2)(3t-10)=0 equation:

(t-2)(3t-10)=0

We multiply parentheses ..

(+3t^2-10t-6t+20)=0

We get rid of parentheses

3t^2-10t-6t+20=0

We add all the numbers together, and all the variables

3t^2-16t+20=0

a = 3; b = -16; c = +20;
Δ = b2-4ac
Δ = -162-4·3·20
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-4}{2*3}=\frac{12}{6}
=2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+4}{2*3}=\frac{20}{6}
=3+1/3
$