x^2+10x=299

Solution for x^2+10x=299 equation:

x^2+10x=299

We move all terms to the left:

x^2+10x-(299)=0

a = 1; b = 10; c = -299;
Δ = b2-4ac
Δ = 102-4·1·(-299)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-36}{2*1}=\frac{-46}{2}
=-23
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+36}{2*1}=\frac{26}{2}
=13
$