2x^2-25x+68=0

Solution for 2x^2-25x+68=0 equation:

2x^2-25x+68=0

a = 2; b = -25; c = +68;
Δ = b2-4ac
Δ = -252-4·2·68
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-9}{2*2}=\frac{16}{4}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+9}{2*2}=\frac{34}{4}
=8+1/2
$