3x2+6x-42=0

Solution for 3x2+6x-42=0 equation:

3x^2+6x-42=0

a = 3; b = 6; c = -42;
Δ = b2-4ac
Δ = 62-4·3·(-42)
Δ = 540
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:

$\sqrt{\Delta}=\sqrt{540}=\sqrt{36*15}=\sqrt{36}*\sqrt{15}=6\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{15}}{2*3}=\frac{-6-6\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{15}}{2*3}=\frac{-6+6\sqrt{15}}{6}
$