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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

-1t^2-t*2+13=0

Wy multiply elements

-1t^2-2t+13=0

a = -1; b = -2; c = +13;
Δ = b2-4ac
Δ = -22-4·(-1)·13
Δ = 56
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:

$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{14}}{2*-1}=\frac{2-2\sqrt{14}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{14}}{2*-1}=\frac{2+2\sqrt{14}}{-2}
$