x^2+0.5x-12=0

Solution for x^2+0.5x-12=0 equation:

x^2+0.5x-12=0

a = 1; b = 0.5; c = -12;
Δ = b2-4ac
Δ = 0.52-4·1·(-12)
Δ = 48.25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0.5)-\sqrt{48.25}}{2*1}=\frac{-0.5-\sqrt{48.25}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0.5)+\sqrt{48.25}}{2*1}=\frac{-0.5+\sqrt{48.25}}{2}
$