7x+3(39x2-5)=x-3

Solution for 7x+3(39x2-5)=x-3 equation:

7x+3(39x^2-5)=x-3

We move all terms to the left:

7x+3(39x^2-5)-(x-3)=0

We multiply parentheses

117x^2+7x-(x-3)-15=0

We get rid of parentheses

117x^2+7x-x+3-15=0

We add all the numbers together, and all the variables

117x^2+6x-12=0

a = 117; b = 6; c = -12;
Δ = b2-4ac
Δ = 62-4·117·(-12)
Δ = 5652
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:

$\sqrt{\Delta}=\sqrt{5652}=\sqrt{36*157}=\sqrt{36}*\sqrt{157}=6\sqrt{157}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{157}}{2*117}=\frac{-6-6\sqrt{157}}{234}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{157}}{2*117}=\frac{-6+6\sqrt{157}}{234}
$