3x^2+24x+34=0

Solution for 3x^2+24x+34=0 equation:

3x^2+24x+34=0

a = 3; b = 24; c = +34;
Δ = b2-4ac
Δ = 242-4·3·34
Δ = 168
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:

$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{42}}{2*3}=\frac{-24-2\sqrt{42}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{42}}{2*3}=\frac{-24+2\sqrt{42}}{6}
$