13-(t+(t^2)/2)=0

Solution for 13-(t+(t^2)/2)=0 equation:

13-(t+(t^2)/2)=0

We get rid of parentheses

-t^2/2-t+13=0

We multiply all the terms by the denominator


-t^2-t*2+13*2=0

We add all the numbers together, and all the variables

-1t^2-t*2+26=0

Wy multiply elements

-1t^2-2t+26=0

a = -1; b = -2; c = +26;
Δ = b2-4ac
Δ = -22-4·(-1)·26
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6\sqrt{3}}{2*-1}=\frac{2-6\sqrt{3}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6\sqrt{3}}{2*-1}=\frac{2+6\sqrt{3}}{-2}
$