2x^2+12x-41=0

Solution for 2x^2+12x-41=0 equation:

2x^2+12x-41=0

a = 2; b = 12; c = -41;
Δ = b2-4ac
Δ = 122-4·2·(-41)
Δ = 472
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:

$\sqrt{\Delta}=\sqrt{472}=\sqrt{4*118}=\sqrt{4}*\sqrt{118}=2\sqrt{118}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{118}}{2*2}=\frac{-12-2\sqrt{118}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{118}}{2*2}=\frac{-12+2\sqrt{118}}{4}
$