x^2+10x=47

Solution for x^2+10x=47 equation:

x^2+10x=47

We move all terms to the left:

x^2+10x-(47)=0

a = 1; b = 10; c = -47;
Δ = b2-4ac
Δ = 102-4·1·(-47)
Δ = 288
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:

$\sqrt{\Delta}=\sqrt{288}=\sqrt{144*2}=\sqrt{144}*\sqrt{2}=12\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-12\sqrt{2}}{2*1}=\frac{-10-12\sqrt{2}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+12\sqrt{2}}{2*1}=\frac{-10+12\sqrt{2}}{2}
$