15(2x-26)=6(3x-29);x

Solution for 15(2x-26)=6(3x-29);x equation:

15(2x-26)=6(3x-29)x

We move all terms to the left:

15(2x-26)-(6(3x-29)x)=0

We multiply parentheses

30x-(6(3x-29)x)-390=0


We calculate terms in parentheses: -(6(3x-29)x), so:

6(3x-29)x

We multiply parentheses

18x^2-174x

Back to the equation:

-(18x^2-174x)


We get rid of parentheses

-18x^2+30x+174x-390=0

We add all the numbers together, and all the variables

-18x^2+204x-390=0

a = -18; b = 204; c = -390;
Δ = b2-4ac
Δ = 2042-4·(-18)·(-390)
Δ = 13536
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:

$\sqrt{\Delta}=\sqrt{13536}=\sqrt{144*94}=\sqrt{144}*\sqrt{94}=12\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(204)-12\sqrt{94}}{2*-18}=\frac{-204-12\sqrt{94}}{-36}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(204)+12\sqrt{94}}{2*-18}=\frac{-204+12\sqrt{94}}{-36}
$