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

Solution for (-3t+5)(2-t)=0 equation:

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

We add all the numbers together, and all the variables

(-3t+5)(-1t+2)=0

We multiply parentheses ..

(+3t^2-6t-5t+10)=0

We get rid of parentheses

3t^2-6t-5t+10=0

We add all the numbers together, and all the variables

3t^2-11t+10=0

a = 3; b = -11; c = +10;
Δ = b2-4ac
Δ = -112-4·3·10
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-1}{2*3}=\frac{10}{6}
=1+2/3
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+1}{2*3}=\frac{12}{6}
=2
$