4t^2-30t+45=0

Solution for 4t^2-30t+45=0 equation:

4t^2-30t+45=0

a = 4; b = -30; c = +45;
Δ = b2-4ac
Δ = -302-4·4·45
Δ = 180
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:

$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-30)-6\sqrt{5}}{2*4}=\frac{30-6\sqrt{5}}{8}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-30)+6\sqrt{5}}{2*4}=\frac{30+6\sqrt{5}}{8}
$