12x^2+x-38=0

Solution for 12x^2+x-38=0 equation:

12x^2+x-38=0

a = 12; b = 1; c = -38;
Δ = b2-4ac
Δ = 12-4·12·(-38)
Δ = 1825
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:

$\sqrt{\Delta}=\sqrt{1825}=\sqrt{25*73}=\sqrt{25}*\sqrt{73}=5\sqrt{73}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-5\sqrt{73}}{2*12}=\frac{-1-5\sqrt{73}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+5\sqrt{73}}{2*12}=\frac{-1+5\sqrt{73}}{24}
$