6(3x+5)=7(2x+2)18x-14

Solution for 6(3x+5)=7(2x+2)18x-14 equation:

6(3x+5)=7(2x+2)18x-14

We move all terms to the left:

6(3x+5)-(7(2x+2)18x-14)=0

We multiply parentheses

18x-(7(2x+2)18x-14)+30=0


We calculate terms in parentheses: -(7(2x+2)18x-14), so:

7(2x+2)18x-14

We multiply parentheses

252x^2+252x-14

Back to the equation:

-(252x^2+252x-14)


We get rid of parentheses

-252x^2+18x-252x+14+30=0

We add all the numbers together, and all the variables

-252x^2-234x+44=0

a = -252; b = -234; c = +44;
Δ = b2-4ac
Δ = -2342-4·(-252)·44
Δ = 99108
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:

$\sqrt{\Delta}=\sqrt{99108}=\sqrt{36*2753}=\sqrt{36}*\sqrt{2753}=6\sqrt{2753}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-234)-6\sqrt{2753}}{2*-252}=\frac{234-6\sqrt{2753}}{-504}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-234)+6\sqrt{2753}}{2*-252}=\frac{234+6\sqrt{2753}}{-504}
$