3(x+4)2+6x=3x2+7

Solution for 3(x+4)2+6x=3x2+7 equation:

3(x+4)2+6x=3x^2+7

We move all terms to the left:

3(x+4)2+6x-(3x^2+7)=0

We add all the numbers together, and all the variables

6x+3(x+4)2-(3x^2+7)=0

We multiply parentheses

6x+6x-(3x^2+7)+24=0

We get rid of parentheses

-3x^2+6x+6x-7+24=0

We add all the numbers together, and all the variables

-3x^2+12x+17=0

a = -3; b = 12; c = +17;
Δ = b2-4ac
Δ = 122-4·(-3)·17
Δ = 348
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:

$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{87}}{2*-3}=\frac{-12-2\sqrt{87}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{87}}{2*-3}=\frac{-12+2\sqrt{87}}{-6}
$