X^2+20x=61

Solution for X^2+20x=61 equation:

X^2+20X=61

We move all terms to the left:

X^2+20X-(61)=0

a = 1; b = 20; c = -61;
Δ = b2-4ac
Δ = 202-4·1·(-61)
Δ = 644
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:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{161}}{2*1}=\frac{-20-2\sqrt{161}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{161}}{2*1}=\frac{-20+2\sqrt{161}}{2}
$