X^2+20x+50=5x-4

Solution for X^2+20x+50=5x-4 equation:

X^2+20X+50=5X-4

We move all terms to the left:

X^2+20X+50-(5X-4)=0

We get rid of parentheses

X^2+20X-5X+4+50=0

We add all the numbers together, and all the variables

X^2+15X+54=0

a = 1; b = 15; c = +54;
Δ = b2-4ac
Δ = 152-4·1·54
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3}{2*1}=\frac{-18}{2}
=-9
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3}{2*1}=\frac{-12}{2}
=-6
$