t^2-30t+50=0

Solution for t^2-30t+50=0 equation:

t^2-30t+50=0

a = 1; b = -30; c = +50;
Δ = b2-4ac
Δ = -302-4·1·50
Δ = 700
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:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-10\sqrt{7}}{2*1}=\frac{30-10\sqrt{7}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+10\sqrt{7}}{2*1}=\frac{30+10\sqrt{7}}{2}
$