-9t^2+180t-7=0

Solution for -9t^2+180t-7=0 equation:

-9t^2+180t-7=0

a = -9; b = 180; c = -7;
Δ = b2-4ac
Δ = 1802-4·(-9)·(-7)
Δ = 32148
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:

$\sqrt{\Delta}=\sqrt{32148}=\sqrt{36*893}=\sqrt{36}*\sqrt{893}=6\sqrt{893}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-6\sqrt{893}}{2*-9}=\frac{-180-6\sqrt{893}}{-18}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+6\sqrt{893}}{2*-9}=\frac{-180+6\sqrt{893}}{-18}
$