3x^2-5(2x^2-1)+4x^2=2

Solution for 3x^2-5(2x^2-1)+4x^2=2 equation:

3x^2-5(2x^2-1)+4x^2=2

We move all terms to the left:

3x^2-5(2x^2-1)+4x^2-(2)=0

We add all the numbers together, and all the variables

7x^2-5(2x^2-1)-2=0

We multiply parentheses

7x^2-10x^2+5-2=0

We add all the numbers together, and all the variables

-3x^2+3=0

a = -3; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-3)·3
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-3}=\frac{-6}{-6}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-3}=\frac{6}{-6}
=-1
$