3x^2+42x-3=0

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

3x^2+42x-3=0

a = 3; b = 42; c = -3;
Δ = b2-4ac
Δ = 422-4·3·(-3)
Δ = 1800
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:

$\sqrt{\Delta}=\sqrt{1800}=\sqrt{900*2}=\sqrt{900}*\sqrt{2}=30\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-30\sqrt{2}}{2*3}=\frac{-42-30\sqrt{2}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+30\sqrt{2}}{2*3}=\frac{-42+30\sqrt{2}}{6}
$