5t^2+12t-80=0

Solution for 5t^2+12t-80=0 equation:

5t^2+12t-80=0

a = 5; b = 12; c = -80;
Δ = b2-4ac
Δ = 122-4·5·(-80)
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-4\sqrt{109}}{2*5}=\frac{-12-4\sqrt{109}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+4\sqrt{109}}{2*5}=\frac{-12+4\sqrt{109}}{10}
$