4x2+6=3x2+18

Solution for 4x2+6=3x2+18 equation:

4x^2+6=3x^2+18

We move all terms to the left:

4x^2+6-(3x^2+18)=0

We get rid of parentheses

4x^2-3x^2-18+6=0

We add all the numbers together, and all the variables

x^2-12=0

a = 1; b = 0; c = -12;
Δ = b2-4ac
Δ = 02-4·1·(-12)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*1}=\frac{0-4\sqrt{3}}{2}
=-\frac{4\sqrt{3}}{2}
=-2\sqrt{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*1}=\frac{0+4\sqrt{3}}{2}
=\frac{4\sqrt{3}}{2}
=2\sqrt{3}
$