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

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

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

We move all terms to the left:

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

We multiply parentheses

3x^2-(4(2x+3))+9=0


We calculate terms in parentheses: -(4(2x+3)), so:

4(2x+3)

We multiply parentheses

8x+12

Back to the equation:

-(8x+12)


We get rid of parentheses

3x^2-8x-12+9=0

We add all the numbers together, and all the variables

3x^2-8x-3=0

a = 3; b = -8; c = -3;
Δ = b2-4ac
Δ = -82-4·3·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-10}{2*3}=\frac{-2}{6}
=-1/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+10}{2*3}=\frac{18}{6}
=3
$