-5t^2+40t-70=0

Solution for -5t^2+40t-70=0 equation:

-5t^2+40t-70=0

a = -5; b = 40; c = -70;
Δ = b2-4ac
Δ = 402-4·(-5)·(-70)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-10\sqrt{2}}{2*-5}=\frac{-40-10\sqrt{2}}{-10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+10\sqrt{2}}{2*-5}=\frac{-40+10\sqrt{2}}{-10}
$