2x^2+16x-47=0

Solution for 2x^2+16x-47=0 equation:

2x^2+16x-47=0

a = 2; b = 16; c = -47;
Δ = b2-4ac
Δ = 162-4·2·(-47)
Δ = 632
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:

$\sqrt{\Delta}=\sqrt{632}=\sqrt{4*158}=\sqrt{4}*\sqrt{158}=2\sqrt{158}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{158}}{2*2}=\frac{-16-2\sqrt{158}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{158}}{2*2}=\frac{-16+2\sqrt{158}}{4}
$