3.25t^2+10t-10=0

Solution for 3.25t^2+10t-10=0 equation:

3.25t^2+10t-10=0

a = 3.25; b = 10; c = -10;
Δ = b2-4ac
Δ = 102-4·3.25·(-10)
Δ = 230
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-\sqrt{230}}{2*3.25}=\frac{-10-\sqrt{230}}{6.5}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+\sqrt{230}}{2*3.25}=\frac{-10+\sqrt{230}}{6.5}
$