t+(t-10)+t2=66

Solution for t+(t-10)+t2=66 equation:

t+(t-10)+t2=66

We move all terms to the left:

t+(t-10)+t2-(66)=0

We add all the numbers together, and all the variables

t^2+t+(t-10)-66=0

We get rid of parentheses

t^2+t+t-10-66=0

We add all the numbers together, and all the variables

t^2+2t-76=0

a = 1; b = 2; c = -76;
Δ = b2-4ac
Δ = 22-4·1·(-76)
Δ = 308
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:

$\sqrt{\Delta}=\sqrt{308}=\sqrt{4*77}=\sqrt{4}*\sqrt{77}=2\sqrt{77}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{77}}{2*1}=\frac{-2-2\sqrt{77}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{77}}{2*1}=\frac{-2+2\sqrt{77}}{2}
$