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

Solution for 2(2x2-5)=3(2-3x)2 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 add all the numbers together, and all the variables

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

We multiply parentheses

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


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

3(-3x+2)2

We multiply parentheses

-18x+12

Back to the equation:

-(-18x+12)


We get rid of parentheses

4x^2+18x-12-10=0

We add all the numbers together, and all the variables

4x^2+18x-22=0

a = 4; b = 18; c = -22;
Δ = b2-4ac
Δ = 182-4·4·(-22)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-26}{2*4}=\frac{-44}{8}
=-5+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+26}{2*4}=\frac{8}{8}
=1
$