2(2x2-5)=3(2-3x2)

Solution for 2(2x2-5)=3(2-3x2) equation:

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

We move all terms to the left:

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

We multiply parentheses

-(3(2-3x^2))+4x^2-10=0


We calculate terms in parentheses: -(3(2-3x^2)), so:

3(2-3x^2)

We multiply parentheses

-9x^2+6

Back to the equation:

-(-9x^2+6)


We add all the numbers together, and all the variables

4x^2-(-9x^2+6)-10=0

We get rid of parentheses

4x^2+9x^2-6-10=0

We add all the numbers together, and all the variables

13x^2-16=0

a = 13; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·13·(-16)
Δ = 832
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:

$\sqrt{\Delta}=\sqrt{832}=\sqrt{64*13}=\sqrt{64}*\sqrt{13}=8\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{13}}{2*13}=\frac{0-8\sqrt{13}}{26}
=-\frac{8\sqrt{13}}{26}
=-\frac{4\sqrt{13}}{13}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{13}}{2*13}=\frac{0+8\sqrt{13}}{26}
=\frac{8\sqrt{13}}{26}
=\frac{4\sqrt{13}}{13}
$