x=3(x^2-4)+7(x-2)

Solution for x=3(x^2-4)+7(x-2) equation:

x=3(x^2-4)+7(x-2)

We move all terms to the left:

x-(3(x^2-4)+7(x-2))=0


We calculate terms in parentheses: -(3(x^2-4)+7(x-2)), so:

3(x^2-4)+7(x-2)

We multiply parentheses

3x^2+7x-12-14

We add all the numbers together, and all the variables

3x^2+7x-26

Back to the equation:

-(3x^2+7x-26)


We get rid of parentheses

-3x^2+x-7x+26=0

We add all the numbers together, and all the variables

-3x^2-6x+26=0

a = -3; b = -6; c = +26;
Δ = b2-4ac
Δ = -62-4·(-3)·26
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{87}}{2*-3}=\frac{6-2\sqrt{87}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{87}}{2*-3}=\frac{6+2\sqrt{87}}{-6}
$