t^2=10t+50

Solution for t^2=10t+50 equation:

t^2=10t+50

We move all terms to the left:

t^2-(10t+50)=0

We get rid of parentheses

t^2-10t-50=0

a = 1; b = -10; c = -50;
Δ = b2-4ac
Δ = -102-4·1·(-50)
Δ = 300
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:

$\sqrt{\Delta}=\sqrt{300}=\sqrt{100*3}=\sqrt{100}*\sqrt{3}=10\sqrt{3}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-10\sqrt{3}}{2*1}=\frac{10-10\sqrt{3}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+10\sqrt{3}}{2*1}=\frac{10+10\sqrt{3}}{2}
$