5t^2+340t+500=0

Solution for 5t^2+340t+500=0 equation:

5t^2+340t+500=0

a = 5; b = 340; c = +500;
Δ = b2-4ac
Δ = 3402-4·5·500
Δ = 105600
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:

$\sqrt{\Delta}=\sqrt{105600}=\sqrt{1600*66}=\sqrt{1600}*\sqrt{66}=40\sqrt{66}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(340)-40\sqrt{66}}{2*5}=\frac{-340-40\sqrt{66}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(340)+40\sqrt{66}}{2*5}=\frac{-340+40\sqrt{66}}{10}
$