2x^2+24x+47=0

Solution for 2x^2+24x+47=0 equation:

2x^2+24x+47=0

a = 2; b = 24; c = +47;
Δ = b2-4ac
Δ = 242-4·2·47
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-10\sqrt{2}}{2*2}=\frac{-24-10\sqrt{2}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+10\sqrt{2}}{2*2}=\frac{-24+10\sqrt{2}}{4}
$