3(2x2-3)=4x2-5

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Solution for 3(2x2-3)=4x2-5 equation:



3(2x^2-3)=4x^2-5
We move all terms to the left:
3(2x^2-3)-(4x^2-5)=0
We multiply parentheses
6x^2-(4x^2-5)-9=0
We get rid of parentheses
6x^2-4x^2+5-9=0
We add all the numbers together, and all the variables
2x^2-4=0
a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
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:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4} =-\frac{4\sqrt{2}}{4} =-\sqrt{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4} =\frac{4\sqrt{2}}{4} =\sqrt{2} $

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