t^2-9t=20=0

Solution for t^2-9t=20=0 equation:

t^2-9t=20=0

We move all terms to the left:

t^2-9t-(20)=0

a = 1; b = -9; c = -20;
Δ = b2-4ac
Δ = -92-4·1·(-20)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{161}}{2*1}=\frac{9-\sqrt{161}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{161}}{2*1}=\frac{9+\sqrt{161}}{2}
$