t^2+10t+4=0

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

t^2+10t+4=0

a = 1; b = 10; c = +4;
Δ = b2-4ac
Δ = 102-4·1·4
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{21}}{2*1}=\frac{-10-2\sqrt{21}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{21}}{2*1}=\frac{-10+2\sqrt{21}}{2}
$