X^2-4x-5x=20

Solution for X^2-4x-5x=20 equation:

X^2-4X-5X=20

We move all terms to the left:

X^2-4X-5X-(20)=0

We add all the numbers together, and all the variables

X^2-9X-20=0

a = 1; b = -9; c = -20;
Δ = b2-4ac
Δ = -92-4·1·(-20)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{161}}{2*1}=\frac{9-\sqrt{161}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{161}}{2*1}=\frac{9+\sqrt{161}}{2}
$