a2+6a-3=0

Solution for a2+6a-3=0 equation:

a2+6a-3=0

We add all the numbers together, and all the variables

a^2+6a-3=0

a = 1; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·1·(-3)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*1}=\frac{-6-4\sqrt{3}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*1}=\frac{-6+4\sqrt{3}}{2}
$