-8t^2+45t-50=0

Solution for -8t^2+45t-50=0 equation:

-8t^2+45t-50=0

a = -8; b = 45; c = -50;
Δ = b2-4ac
Δ = 452-4·(-8)·(-50)
Δ = 425
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:

$\sqrt{\Delta}=\sqrt{425}=\sqrt{25*17}=\sqrt{25}*\sqrt{17}=5\sqrt{17}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(45)-5\sqrt{17}}{2*-8}=\frac{-45-5\sqrt{17}}{-16}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(45)+5\sqrt{17}}{2*-8}=\frac{-45+5\sqrt{17}}{-16}
$