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

Solution for x(2x-3)=(2x+8x)(x-5) equation:

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

We move all terms to the left:

x(2x-3)-((2x+8x)(x-5))=0

We add all the numbers together, and all the variables

x(2x-3)-((+10x)(x-5))=0

We multiply parentheses

2x^2-3x-((+10x)(x-5))=0

We multiply parentheses ..

2x^2-((+10x^2-50x))-3x=0


We calculate terms in parentheses: -((+10x^2-50x)), so:

(+10x^2-50x)

We get rid of parentheses

10x^2-50x

Back to the equation:

-(10x^2-50x)


We add all the numbers together, and all the variables

2x^2-3x-(10x^2-50x)=0

We get rid of parentheses

2x^2-10x^2-3x+50x=0

We add all the numbers together, and all the variables

-8x^2+47x=0

a = -8; b = 47; c = 0;
Δ = b2-4ac
Δ = 472-4·(-8)·0
Δ = 2209
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2209}=47$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(47)-47}{2*-8}=\frac{-94}{-16}
=5+7/8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(47)+47}{2*-8}=\frac{0}{-16}
=0
$