t^2-30t+25=0

Solution for t^2-30t+25=0 equation:

t^2-30t+25=0

a = 1; b = -30; c = +25;
Δ = b2-4ac
Δ = -302-4·1·25
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-20\sqrt{2}}{2*1}=\frac{30-20\sqrt{2}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+20\sqrt{2}}{2*1}=\frac{30+20\sqrt{2}}{2}
$