5t^2+96t-172=0

Solution for 5t^2+96t-172=0 equation:

5t^2+96t-172=0

a = 5; b = 96; c = -172;
Δ = b2-4ac
Δ = 962-4·5·(-172)
Δ = 12656
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:

$\sqrt{\Delta}=\sqrt{12656}=\sqrt{16*791}=\sqrt{16}*\sqrt{791}=4\sqrt{791}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-4\sqrt{791}}{2*5}=\frac{-96-4\sqrt{791}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+4\sqrt{791}}{2*5}=\frac{-96+4\sqrt{791}}{10}
$