20x^2+48x-5=(2x+5)

Solution for 20x^2+48x-5=(2x+5) equation:

20x^2+48x-5=(2x+5)

We move all terms to the left:

20x^2+48x-5-((2x+5))=0


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

(2x+5)

We get rid of parentheses

2x+5

Back to the equation:

-(2x+5)


We get rid of parentheses

20x^2+48x-2x-5-5=0

We add all the numbers together, and all the variables

20x^2+46x-10=0

a = 20; b = 46; c = -10;
Δ = b2-4ac
Δ = 462-4·20·(-10)
Δ = 2916
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2916}=54$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(46)-54}{2*20}=\frac{-100}{40}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(46)+54}{2*20}=\frac{8}{40}
=1/5
$