a2+4a-500=0

Solution for a2+4a-500=0 equation:

a2+4a-500=0

We add all the numbers together, and all the variables

a^2+4a-500=0

a = 1; b = 4; c = -500;
Δ = b2-4ac
Δ = 42-4·1·(-500)
Δ = 2016
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2016}=\sqrt{144*14}=\sqrt{144}*\sqrt{14}=12\sqrt{14}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12\sqrt{14}}{2*1}=\frac{-4-12\sqrt{14}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12\sqrt{14}}{2*1}=\frac{-4+12\sqrt{14}}{2}
$