3x2+42x+105=0

Solution for 3x2+42x+105=0 equation:

3x^2+42x+105=0

a = 3; b = 42; c = +105;
Δ = b2-4ac
Δ = 422-4·3·105
Δ = 504
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:

$\sqrt{\Delta}=\sqrt{504}=\sqrt{36*14}=\sqrt{36}*\sqrt{14}=6\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{14}}{2*3}=\frac{-42-6\sqrt{14}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{14}}{2*3}=\frac{-42+6\sqrt{14}}{6}
$