8x^2-42x+40=0

Solution for 8x^2-42x+40=0 equation:

8x^2-42x+40=0

a = 8; b = -42; c = +40;
Δ = b2-4ac
Δ = -422-4·8·40
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-22}{2*8}=\frac{20}{16}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+22}{2*8}=\frac{64}{16}
=4
$