(4x^2-41x+40)=(x-9)

Solution for (4x^2-41x+40)=(x-9) equation:

(4x^2-41x+40)=(x-9)

We move all terms to the left:

(4x^2-41x+40)-((x-9))=0

We get rid of parentheses

4x^2-41x-((x-9))+40=0


We calculate terms in parentheses: -((x-9)), so:

(x-9)

We get rid of parentheses

x-9

Back to the equation:

-(x-9)


We get rid of parentheses

4x^2-41x-x+9+40=0

We add all the numbers together, and all the variables

4x^2-42x+49=0

a = 4; b = -42; c = +49;
Δ = b2-4ac
Δ = -422-4·4·49
Δ = 980
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:

$\sqrt{\Delta}=\sqrt{980}=\sqrt{196*5}=\sqrt{196}*\sqrt{5}=14\sqrt{5}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-42)-14\sqrt{5}}{2*4}=\frac{42-14\sqrt{5}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-42)+14\sqrt{5}}{2*4}=\frac{42+14\sqrt{5}}{8}
$