t2-10t+9=0

Solution for t2-10t+9=0 equation:

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