2t^2-19=t+1-t^2=

Solution for 2t^2-19=t+1-t^2= equation:

2t^2-19=t+1-t^2=

We move all terms to the left:

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

3t^2-1t-20=0

a = 3; b = -1; c = -20;
Δ = b2-4ac
Δ = -12-4·3·(-20)
Δ = 241
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{-(-1)-\sqrt{241}}{2*3}=\frac{1-\sqrt{241}}{6}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{241}}{2*3}=\frac{1+\sqrt{241}}{6}
$