2x^2+22x+47=0

Solution for 2x^2+22x+47=0 equation:

2x^2+22x+47=0

a = 2; b = 22; c = +47;
Δ = b2-4ac
Δ = 222-4·2·47
Δ = 108
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:

$\sqrt{\Delta}=\sqrt{108}=\sqrt{36*3}=\sqrt{36}*\sqrt{3}=6\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-6\sqrt{3}}{2*2}=\frac{-22-6\sqrt{3}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+6\sqrt{3}}{2*2}=\frac{-22+6\sqrt{3}}{4}
$