3(x^2-2)=4(x+1)

Solution for 3(x^2-2)=4(x+1) equation:

3(x^2-2)=4(x+1)

We move all terms to the left:

3(x^2-2)-(4(x+1))=0

We multiply parentheses

3x^2-(4(x+1))-6=0


We calculate terms in parentheses: -(4(x+1)), so:

4(x+1)

We multiply parentheses

4x+4

Back to the equation:

-(4x+4)


We get rid of parentheses

3x^2-4x-4-6=0

We add all the numbers together, and all the variables

3x^2-4x-10=0

a = 3; b = -4; c = -10;
Δ = b2-4ac
Δ = -42-4·3·(-10)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{34}}{2*3}=\frac{4-2\sqrt{34}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{34}}{2*3}=\frac{4+2\sqrt{34}}{6}
$