3x2+69x+69=0

Solution for 3x2+69x+69=0 equation:

3x^2+69x+69=0

a = 3; b = 69; c = +69;
Δ = b2-4ac
Δ = 692-4·3·69
Δ = 3933
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:

$\sqrt{\Delta}=\sqrt{3933}=\sqrt{9*437}=\sqrt{9}*\sqrt{437}=3\sqrt{437}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(69)-3\sqrt{437}}{2*3}=\frac{-69-3\sqrt{437}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(69)+3\sqrt{437}}{2*3}=\frac{-69+3\sqrt{437}}{6}
$