(3x2+4)-10=12(4x+16)

Solution for (3x2+4)-10=12(4x+16) equation:

(3x^2+4)-10=12(4x+16)

We move all terms to the left:

(3x^2+4)-10-(12(4x+16))=0

We get rid of parentheses

3x^2-(12(4x+16))+4-10=0


We calculate terms in parentheses: -(12(4x+16)), so:

12(4x+16)

We multiply parentheses

48x+192

Back to the equation:

-(48x+192)


We add all the numbers together, and all the variables

3x^2-(48x+192)-6=0

We get rid of parentheses

3x^2-48x-192-6=0

We add all the numbers together, and all the variables

3x^2-48x-198=0

a = 3; b = -48; c = -198;
Δ = b2-4ac
Δ = -482-4·3·(-198)
Δ = 4680
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:

$\sqrt{\Delta}=\sqrt{4680}=\sqrt{36*130}=\sqrt{36}*\sqrt{130}=6\sqrt{130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-48)-6\sqrt{130}}{2*3}=\frac{48-6\sqrt{130}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-48)+6\sqrt{130}}{2*3}=\frac{48+6\sqrt{130}}{6}
$