0=-2t^2+7t-4

Solution for 0=-2t^2+7t-4 equation:

0=-2t^2+7t-4

We move all terms to the left:

0-(-2t^2+7t-4)=0

We add all the numbers together, and all the variables

-(-2t^2+7t-4)=0

We get rid of parentheses

2t^2-7t+4=0

a = 2; b = -7; c = +4;
Δ = b2-4ac
Δ = -72-4·2·4
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{17}}{2*2}=\frac{7-\sqrt{17}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{17}}{2*2}=\frac{7+\sqrt{17}}{4}
$