20(x+26)=(3x+8)(x-1)

Solution for 20(x+26)=(3x+8)(x-1) equation:

20(x+26)=(3x+8)(x-1)

We move all terms to the left:

20(x+26)-((3x+8)(x-1))=0

We multiply parentheses

20x-((3x+8)(x-1))+520=0

We multiply parentheses ..

-((+3x^2-3x+8x-8))+20x+520=0


We calculate terms in parentheses: -((+3x^2-3x+8x-8)), so:

(+3x^2-3x+8x-8)

We get rid of parentheses

3x^2-3x+8x-8

We add all the numbers together, and all the variables

3x^2+5x-8

Back to the equation:

-(3x^2+5x-8)


We add all the numbers together, and all the variables

20x-(3x^2+5x-8)+520=0

We get rid of parentheses

-3x^2+20x-5x+8+520=0

We add all the numbers together, and all the variables

-3x^2+15x+528=0

a = -3; b = 15; c = +528;
Δ = b2-4ac
Δ = 152-4·(-3)·528
Δ = 6561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6561}=81$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-81}{2*-3}=\frac{-96}{-6}
=+16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+81}{2*-3}=\frac{66}{-6}
=-11
$