x2+20x+61=10

Solution for x2+20x+61=10 equation:

x2+20x+61=10

We move all terms to the left:

x2+20x+61-(10)=0

We add all the numbers together, and all the variables

x^2+20x+51=0

a = 1; b = 20; c = +51;
Δ = b2-4ac
Δ = 202-4·1·51
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-14}{2*1}=\frac{-34}{2}
=-17
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+14}{2*1}=\frac{-6}{2}
=-3
$