22x2-20x+4=(5x-2)

Solution for 22x2-20x+4=(5x-2) equation:

22x^2-20x+4=(5x-2)

We move all terms to the left:

22x^2-20x+4-((5x-2))=0


We calculate terms in parentheses: -((5x-2)), so:

(5x-2)

We get rid of parentheses

5x-2

Back to the equation:

-(5x-2)


We get rid of parentheses

22x^2-20x-5x+2+4=0

We add all the numbers together, and all the variables

22x^2-25x+6=0

a = 22; b = -25; c = +6;
Δ = b2-4ac
Δ = -252-4·22·6
Δ = 97
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{-(-25)-\sqrt{97}}{2*22}=\frac{25-\sqrt{97}}{44}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+\sqrt{97}}{2*22}=\frac{25+\sqrt{97}}{44}
$