(3x^2+4x+6)=(7x^2-1)

Solution for (3x^2+4x+6)=(7x^2-1) equation:

(3x^2+4x+6)=(7x^2-1)

We move all terms to the left:

(3x^2+4x+6)-((7x^2-1))=0

We get rid of parentheses

3x^2+4x-((7x^2-1))+6=0


We calculate terms in parentheses: -((7x^2-1)), so:

(7x^2-1)

We get rid of parentheses

7x^2-1

Back to the equation:

-(7x^2-1)


We get rid of parentheses

3x^2-7x^2+4x+1+6=0

We add all the numbers together, and all the variables

-4x^2+4x+7=0

a = -4; b = 4; c = +7;
Δ = b2-4ac
Δ = 42-4·(-4)·7
Δ = 128
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:

$\sqrt{\Delta}=\sqrt{128}=\sqrt{64*2}=\sqrt{64}*\sqrt{2}=8\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{2}}{2*-4}=\frac{-4-8\sqrt{2}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{2}}{2*-4}=\frac{-4+8\sqrt{2}}{-8}
$