1.25t^2+10t-30=0

Solution for 1.25t^2+10t-30=0 equation:

1.25t^2+10t-30=0

a = 1.25; b = 10; c = -30;
Δ = b2-4ac
Δ = 102-4·1.25·(-30)
Δ = 250
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:

$\sqrt{\Delta}=\sqrt{250}=\sqrt{25*10}=\sqrt{25}*\sqrt{10}=5\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-5\sqrt{10}}{2*1.25}=\frac{-10-5\sqrt{10}}{2.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+5\sqrt{10}}{2*1.25}=\frac{-10+5\sqrt{10}}{2.5}
$