40x^2+40x+3=0

Solution for 40x^2+40x+3=0 equation:

40x^2+40x+3=0

a = 40; b = 40; c = +3;
Δ = b2-4ac
Δ = 402-4·40·3
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-4\sqrt{70}}{2*40}=\frac{-40-4\sqrt{70}}{80}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+4\sqrt{70}}{2*40}=\frac{-40+4\sqrt{70}}{80}
$