a^2+10a=5

Solution for a^2+10a=5 equation:

a^2+10a=5

We move all terms to the left:

a^2+10a-(5)=0

a = 1; b = 10; c = -5;
Δ = b2-4ac
Δ = 102-4·1·(-5)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{30}}{2*1}=\frac{-10-2\sqrt{30}}{2}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{30}}{2*1}=\frac{-10+2\sqrt{30}}{2}
$