10x^2-42x+40=0

Solution for 10x^2-42x+40=0 equation:

10x^2-42x+40=0

a = 10; b = -42; c = +40;
Δ = b2-4ac
Δ = -422-4·10·40
Δ = 164
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:

$\sqrt{\Delta}=\sqrt{164}=\sqrt{4*41}=\sqrt{4}*\sqrt{41}=2\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{41}}{2*10}=\frac{42-2\sqrt{41}}{20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{41}}{2*10}=\frac{42+2\sqrt{41}}{20}
$