4(2x2+5)=3(2x2+8)

Solution for 4(2x2+5)=3(2x2+8) equation:

4(2x^2+5)=3(2x^2+8)

We move all terms to the left:

4(2x^2+5)-(3(2x^2+8))=0

We multiply parentheses

8x^2-(3(2x^2+8))+20=0


We calculate terms in parentheses: -(3(2x^2+8)), so:

3(2x^2+8)

We multiply parentheses

6x^2+24

Back to the equation:

-(6x^2+24)


We get rid of parentheses

8x^2-6x^2-24+20=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}
$