4t^2+10t+5=0

Solution for 4t^2+10t+5=0 equation:

4t^2+10t+5=0

a = 4; b = 10; c = +5;
Δ = b2-4ac
Δ = 102-4·4·5
Δ = 20
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:

$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{5}}{2*4}=\frac{-10-2\sqrt{5}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{5}}{2*4}=\frac{-10+2\sqrt{5}}{8}
$