2x^2+42x-3=0

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

2x^2+42x-3=0

a = 2; b = 42; c = -3;
Δ = b2-4ac
Δ = 422-4·2·(-3)
Δ = 1788
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:

$\sqrt{\Delta}=\sqrt{1788}=\sqrt{4*447}=\sqrt{4}*\sqrt{447}=2\sqrt{447}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-2\sqrt{447}}{2*2}=\frac{-42-2\sqrt{447}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+2\sqrt{447}}{2*2}=\frac{-42+2\sqrt{447}}{4}
$