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

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

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

We move all terms to the left:

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

We multiply parentheses

-4x^2-(2(4-x^2)+2)+6x^2-15=0


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

2(4-x^2)+2

We multiply parentheses

-2x^2+8+2

We add all the numbers together, and all the variables

-2x^2+10

Back to the equation:

-(-2x^2+10)


We add all the numbers together, and all the variables

2x^2-(-2x^2+10)-15=0

We get rid of parentheses

2x^2+2x^2-10-15=0

We add all the numbers together, and all the variables

4x^2-25=0

a = 4; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·4·(-25)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20}{2*4}=\frac{-20}{8}
=-2+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20}{2*4}=\frac{20}{8}
=2+1/2
$