0=t^2+8t+10

Solution for 0=t^2+8t+10 equation:

0=t^2+8t+10

We move all terms to the left:

0-(t^2+8t+10)=0

We add all the numbers together, and all the variables

-(t^2+8t+10)=0

We get rid of parentheses

-t^2-8t-10=0

We add all the numbers together, and all the variables

-1t^2-8t-10=0

a = -1; b = -8; c = -10;
Δ = b2-4ac
Δ = -82-4·(-1)·(-10)
Δ = 24
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:

$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-2\sqrt{6}}{2*-1}=\frac{8-2\sqrt{6}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+2\sqrt{6}}{2*-1}=\frac{8+2\sqrt{6}}{-2}
$