2t^2+19t-9=5-t

Solution for 2t^2+19t-9=5-t equation:

2t^2+19t-9=5-t

We move all terms to the left:

2t^2+19t-9-(5-t)=0

We add all the numbers together, and all the variables

2t^2+19t-(-1t+5)-9=0

We get rid of parentheses

2t^2+19t+1t-5-9=0

We add all the numbers together, and all the variables

2t^2+20t-14=0

a = 2; b = 20; c = -14;
Δ = b2-4ac
Δ = 202-4·2·(-14)
Δ = 512
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{512}=\sqrt{256*2}=\sqrt{256}*\sqrt{2}=16\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-16\sqrt{2}}{2*2}=\frac{-20-16\sqrt{2}}{4}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+16\sqrt{2}}{2*2}=\frac{-20+16\sqrt{2}}{4}
$