0.8t^2-24t+90=0

Solution for 0.8t^2-24t+90=0 equation:

0.8t^2-24t+90=0

a = 0.8; b = -24; c = +90;
Δ = b2-4ac
Δ = -242-4·0.8·90
Δ = 288
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-12\sqrt{2}}{2*0.8}=\frac{24-12\sqrt{2}}{1.6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+12\sqrt{2}}{2*0.8}=\frac{24+12\sqrt{2}}{1.6}
$