3x^2-42x-67=0

Solution for 3x^2-42x-67=0 equation:

3x^2-42x-67=0

a = 3; b = -42; c = -67;
Δ = b2-4ac
Δ = -422-4·3·(-67)
Δ = 2568
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:

$\sqrt{\Delta}=\sqrt{2568}=\sqrt{4*642}=\sqrt{4}*\sqrt{642}=2\sqrt{642}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-2\sqrt{642}}{2*3}=\frac{42-2\sqrt{642}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+2\sqrt{642}}{2*3}=\frac{42+2\sqrt{642}}{6}
$