0=t^2+12t-34

Solution for 0=t^2+12t-34 equation:

0=t^2+12t-34

We move all terms to the left:

0-(t^2+12t-34)=0

We add all the numbers together, and all the variables

-(t^2+12t-34)=0

We get rid of parentheses

-t^2-12t+34=0

We add all the numbers together, and all the variables

-1t^2-12t+34=0

a = -1; b = -12; c = +34;
Δ = b2-4ac
Δ = -122-4·(-1)·34
Δ = 280
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:

$\sqrt{\Delta}=\sqrt{280}=\sqrt{4*70}=\sqrt{4}*\sqrt{70}=2\sqrt{70}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-2\sqrt{70}}{2*-1}=\frac{12-2\sqrt{70}}{-2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+2\sqrt{70}}{2*-1}=\frac{12+2\sqrt{70}}{-2}
$