-10t^2+105t-45=0

Solution for -10t^2+105t-45=0 equation:

-10t^2+105t-45=0

a = -10; b = 105; c = -45;
Δ = b2-4ac
Δ = 1052-4·(-10)·(-45)
Δ = 9225
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:

$\sqrt{\Delta}=\sqrt{9225}=\sqrt{225*41}=\sqrt{225}*\sqrt{41}=15\sqrt{41}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(105)-15\sqrt{41}}{2*-10}=\frac{-105-15\sqrt{41}}{-20}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(105)+15\sqrt{41}}{2*-10}=\frac{-105+15\sqrt{41}}{-20}
$