5t^2-45t+25=0

Solution for 5t^2-45t+25=0 equation:

5t^2-45t+25=0

a = 5; b = -45; c = +25;
Δ = b2-4ac
Δ = -452-4·5·25
Δ = 1525
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:

$\sqrt{\Delta}=\sqrt{1525}=\sqrt{25*61}=\sqrt{25}*\sqrt{61}=5\sqrt{61}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-45)-5\sqrt{61}}{2*5}=\frac{45-5\sqrt{61}}{10}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-45)+5\sqrt{61}}{2*5}=\frac{45+5\sqrt{61}}{10}
$