2x^2+42x+9=0

Solution for 2x^2+42x+9=0 equation:

2x^2+42x+9=0

a = 2; b = 42; c = +9;
Δ = b2-4ac
Δ = 422-4·2·9
Δ = 1692
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:

$\sqrt{\Delta}=\sqrt{1692}=\sqrt{36*47}=\sqrt{36}*\sqrt{47}=6\sqrt{47}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(42)-6\sqrt{47}}{2*2}=\frac{-42-6\sqrt{47}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(42)+6\sqrt{47}}{2*2}=\frac{-42+6\sqrt{47}}{4}
$