t^2=10t+10

Solution for t^2=10t+10 equation:

t^2=10t+10

We move all terms to the left:

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

We get rid of parentheses

t^2-10t-10=0

a = 1; b = -10; c = -10;
Δ = b2-4ac
Δ = -102-4·1·(-10)
Δ = 140
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:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*1}=\frac{10-2\sqrt{35}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*1}=\frac{10+2\sqrt{35}}{2}
$