0.9t^2+2t-50=0

Solution for 0.9t^2+2t-50=0 equation:

0.9t^2+2t-50=0

a = 0.9; b = 2; c = -50;
Δ = b2-4ac
Δ = 22-4·0.9·(-50)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{46}}{2*0.9}=\frac{-2-2\sqrt{46}}{1.8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{46}}{2*0.9}=\frac{-2+2\sqrt{46}}{1.8}
$