.4t^2=5t+17.5

Solution for .4t^2=5t+17.5 equation:

.4t^2=5t+17.5

We move all terms to the left:

.4t^2-(5t+17.5)=0

We get rid of parentheses

.4t^2-5t-17.5=0

a = .4; b = -5; c = -17.5;
Δ = b2-4ac
Δ = -52-4·.4·(-17.5)
Δ = 53
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{-(-5)-\sqrt{53}}{2*.4}=\frac{5-\sqrt{53}}{0.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+\sqrt{53}}{2*.4}=\frac{5+\sqrt{53}}{0.8}
$