20x^2+28x-3=(10x-1)

Solution for 20x^2+28x-3=(10x-1) equation:

20x^2+28x-3=(10x-1)

We move all terms to the left:

20x^2+28x-3-((10x-1))=0


We calculate terms in parentheses: -((10x-1)), so:

(10x-1)

We get rid of parentheses

10x-1

Back to the equation:

-(10x-1)


We get rid of parentheses

20x^2+28x-10x+1-3=0

We add all the numbers together, and all the variables

20x^2+18x-2=0

a = 20; b = 18; c = -2;
Δ = b2-4ac
Δ = 182-4·20·(-2)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-22}{2*20}=\frac{-40}{40}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+22}{2*20}=\frac{4}{40}
=1/10
$