5t^2+10t+3=0

Solution for 5t^2+10t+3=0 equation:

5t^2+10t+3=0

a = 5; b = 10; c = +3;
Δ = b2-4ac
Δ = 102-4·5·3
Δ = 40
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:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{10}}{2*5}=\frac{-10-2\sqrt{10}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{10}}{2*5}=\frac{-10+2\sqrt{10}}{10}
$