2x2-12x-38=0

Solution for 2x2-12x-38=0 equation:

2x^2-12x-38=0

a = 2; b = -12; c = -38;
Δ = b2-4ac
Δ = -122-4·2·(-38)
Δ = 448
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:

$\sqrt{\Delta}=\sqrt{448}=\sqrt{64*7}=\sqrt{64}*\sqrt{7}=8\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-8\sqrt{7}}{2*2}=\frac{12-8\sqrt{7}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+8\sqrt{7}}{2*2}=\frac{12+8\sqrt{7}}{4}
$