t2+19t+90=0

Solution for t2+19t+90=0 equation:

t2+19t+90=0

We add all the numbers together, and all the variables

t^2+19t+90=0

a = 1; b = 19; c = +90;
Δ = b2-4ac
Δ = 192-4·1·90
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-1}{2*1}=\frac{-20}{2}
=-10
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+1}{2*1}=\frac{-18}{2}
=-9
$