x^2+16x-133=10

Solution for x^2+16x-133=10 equation:

x^2+16x-133=10

We move all terms to the left:

x^2+16x-133-(10)=0

We add all the numbers together, and all the variables

x^2+16x-143=0

a = 1; b = 16; c = -143;
Δ = b2-4ac
Δ = 162-4·1·(-143)
Δ = 828
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:

$\sqrt{\Delta}=\sqrt{828}=\sqrt{36*23}=\sqrt{36}*\sqrt{23}=6\sqrt{23}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-6\sqrt{23}}{2*1}=\frac{-16-6\sqrt{23}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+6\sqrt{23}}{2*1}=\frac{-16+6\sqrt{23}}{2}
$