x^2+10x=57

Solution for x^2+10x=57 equation:

x^2+10x=57

We move all terms to the left:

x^2+10x-(57)=0

a = 1; b = 10; c = -57;
Δ = b2-4ac
Δ = 102-4·1·(-57)
Δ = 328
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{328}=\sqrt{4*82}=\sqrt{4}*\sqrt{82}=2\sqrt{82}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-2\sqrt{82}}{2*1}=\frac{-10-2\sqrt{82}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{82}}{2*1}=\frac{-10+2\sqrt{82}}{2}
$