5t^2+12t-50=0

Solution for 5t^2+12t-50=0 equation:

5t^2+12t-50=0

a = 5; b = 12; c = -50;
Δ = b2-4ac
Δ = 122-4·5·(-50)
Δ = 1144
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:

$\sqrt{\Delta}=\sqrt{1144}=\sqrt{4*286}=\sqrt{4}*\sqrt{286}=2\sqrt{286}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{286}}{2*5}=\frac{-12-2\sqrt{286}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{286}}{2*5}=\frac{-12+2\sqrt{286}}{10}
$