a^2+10a=600

Solution for a^2+10a=600 equation:

a^2+10a=600

We move all terms to the left:

a^2+10a-(600)=0

a = 1; b = 10; c = -600;
Δ = b2-4ac
Δ = 102-4·1·(-600)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-50}{2*1}=\frac{-60}{2}
=-30
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+50}{2*1}=\frac{40}{2}
=20
$