5x2+40x+8=0

Solution for 5x2+40x+8=0 equation:

5x^2+40x+8=0

a = 5; b = 40; c = +8;
Δ = b2-4ac
Δ = 402-4·5·8
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-12\sqrt{10}}{2*5}=\frac{-40-12\sqrt{10}}{10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+12\sqrt{10}}{2*5}=\frac{-40+12\sqrt{10}}{10}
$