5t2+20t+3=0

Solution for 5t2+20t+3=0 equation:

5t^2+20t+3=0

a = 5; b = 20; c = +3;
Δ = b2-4ac
Δ = 202-4·5·3
Δ = 340
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:

$\sqrt{\Delta}=\sqrt{340}=\sqrt{4*85}=\sqrt{4}*\sqrt{85}=2\sqrt{85}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{85}}{2*5}=\frac{-20-2\sqrt{85}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{85}}{2*5}=\frac{-20+2\sqrt{85}}{10}
$