2t^2=5t-3=0

Solution for 2t^2=5t-3=0 equation:

2t^2=5t-3=0

We move all terms to the left:

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

We get rid of parentheses

2t^2-5t+3=0

a = 2; b = -5; c = +3;
Δ = b2-4ac
Δ = -52-4·2·3
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{-(-5)-1}{2*2}=\frac{4}{4}
=1
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+1}{2*2}=\frac{6}{4}
=1+1/2
$