5(5x-4)-8(x-3)=10x(x+6)

Solution for 5(5x-4)-8(x-3)=10x(x+6) equation:

5(5x-4)-8(x-3)=10x(x+6)

We move all terms to the left:

5(5x-4)-8(x-3)-(10x(x+6))=0

We multiply parentheses

25x-8x-(10x(x+6))-20+24=0


We calculate terms in parentheses: -(10x(x+6)), so:

10x(x+6)

We multiply parentheses

10x^2+60x

Back to the equation:

-(10x^2+60x)


We add all the numbers together, and all the variables

17x-(10x^2+60x)+4=0

We get rid of parentheses

-10x^2+17x-60x+4=0

We add all the numbers together, and all the variables

-10x^2-43x+4=0

a = -10; b = -43; c = +4;
Δ = b2-4ac
Δ = -432-4·(-10)·4
Δ = 2009
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2009}=\sqrt{49*41}=\sqrt{49}*\sqrt{41}=7\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-43)-7\sqrt{41}}{2*-10}=\frac{43-7\sqrt{41}}{-20}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-43)+7\sqrt{41}}{2*-10}=\frac{43+7\sqrt{41}}{-20}
$