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)=0

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

We add all the numbers together, and all the variables

-4x^2+4x+6=0

a = -4; b = 4; c = +6;
Δ = b2-4ac
Δ = 42-4·(-4)·6
Δ = 112
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:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{7}}{2*-4}=\frac{-4-4\sqrt{7}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{7}}{2*-4}=\frac{-4+4\sqrt{7}}{-8}
$