2t^2-25t+77=0

Solution for 2t^2-25t+77=0 equation:

2t^2-25t+77=0

a = 2; b = -25; c = +77;
Δ = b2-4ac
Δ = -252-4·2·77
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-3}{2*2}=\frac{22}{4}
=5+1/2
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+3}{2*2}=\frac{28}{4}
=7
$