t2+10t+16=0

Solution for t2+10t+16=0 equation:

t2+10t+16=0

We add all the numbers together, and all the variables

t^2+10t+16=0

a = 1; b = 10; c = +16;
Δ = b2-4ac
Δ = 102-4·1·16
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*1}=\frac{-16}{2}
=-8
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*1}=\frac{-4}{2}
=-2
$