2x^2-26x+40=0

Solution for 2x^2-26x+40=0 equation:

2x^2-26x+40=0

a = 2; b = -26; c = +40;
Δ = b2-4ac
Δ = -262-4·2·40
Δ = 356
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{356}=\sqrt{4*89}=\sqrt{4}*\sqrt{89}=2\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-26)-2\sqrt{89}}{2*2}=\frac{26-2\sqrt{89}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-26)+2\sqrt{89}}{2*2}=\frac{26+2\sqrt{89}}{4}
$