V=2t^2-5t+3

Solution for V=2t^2-5t+3 equation:

=2V^2-5V+3

We move all terms to the left:

-(2V^2-5V+3)=0

We get rid of parentheses

-2V^2+5V-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:
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$V_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-1}{2*-2}=\frac{-6}{-4}
=1+1/2
$
$V_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+1}{2*-2}=\frac{-4}{-4}
=1
$