(4t-5)(2t-3)=0

Solution for (4t-5)(2t-3)=0 equation:

(4t-5)(2t-3)=0

We multiply parentheses ..

(+8t^2-12t-10t+15)=0

We get rid of parentheses

8t^2-12t-10t+15=0

We add all the numbers together, and all the variables

8t^2-22t+15=0

a = 8; b = -22; c = +15;
Δ = b2-4ac
Δ = -222-4·8·15
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2}{2*8}=\frac{20}{16}
=1+1/4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2}{2*8}=\frac{24}{16}
=1+1/2
$