3a^2+38a+34=0

Solution for 3a^2+38a+34=0 equation:

3a^2+38a+34=0

a = 3; b = 38; c = +34;
Δ = b2-4ac
Δ = 382-4·3·34
Δ = 1036
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1036}=\sqrt{4*259}=\sqrt{4}*\sqrt{259}=2\sqrt{259}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{259}}{2*3}=\frac{-38-2\sqrt{259}}{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{259}}{2*3}=\frac{-38+2\sqrt{259}}{6}
$