x^2+19x=60

Solution for x^2+19x=60 equation:

x^2+19x=60

We move all terms to the left:

x^2+19x-(60)=0

a = 1; b = 19; c = -60;
Δ = b2-4ac
Δ = 192-4·1·(-60)
Δ = 601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-\sqrt{601}}{2*1}=\frac{-19-\sqrt{601}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+\sqrt{601}}{2*1}=\frac{-19+\sqrt{601}}{2}
$