3x^2+42x=377

Solution for 3x^2+42x=377 equation:

3x^2+42x=377

We move all terms to the left:

3x^2+42x-(377)=0

a = 3; b = 42; c = -377;
Δ = b2-4ac
Δ = 422-4·3·(-377)
Δ = 6288
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:

$\sqrt{\Delta}=\sqrt{6288}=\sqrt{16*393}=\sqrt{16}*\sqrt{393}=4\sqrt{393}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-4\sqrt{393}}{2*3}=\frac{-42-4\sqrt{393}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+4\sqrt{393}}{2*3}=\frac{-42+4\sqrt{393}}{6}
$