t2+10t=-9

Solution for t2+10t=-9 equation:

t2+10t=-9

We move all terms to the left:

t2+10t-(-9)=0

We add all the numbers together, and all the variables

t^2+10t+9=0

a = 1; b = 10; c = +9;
Δ = b2-4ac
Δ = 102-4·1·9
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8}{2*1}=\frac{-18}{2}
=-9
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8}{2*1}=\frac{-2}{2}
=-1
$