a^2+16a+3=10

Solution for a^2+16a+3=10 equation:

a^2+16a+3=10

We move all terms to the left:

a^2+16a+3-(10)=0

We add all the numbers together, and all the variables

a^2+16a-7=0

a = 1; b = 16; c = -7;
Δ = b2-4ac
Δ = 162-4·1·(-7)
Δ = 284
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:

$\sqrt{\Delta}=\sqrt{284}=\sqrt{4*71}=\sqrt{4}*\sqrt{71}=2\sqrt{71}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{71}}{2*1}=\frac{-16-2\sqrt{71}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{71}}{2*1}=\frac{-16+2\sqrt{71}}{2}
$