3x^2+4x^2=12x+16

Solution for 3x^2+4x^2=12x+16 equation:

3x^2+4x^2=12x+16

We move all terms to the left:

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

We add all the numbers together, and all the variables

7x^2-(12x+16)=0

We get rid of parentheses

7x^2-12x-16=0

a = 7; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·7·(-16)
Δ = 592
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:

$\sqrt{\Delta}=\sqrt{592}=\sqrt{16*37}=\sqrt{16}*\sqrt{37}=4\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{37}}{2*7}=\frac{12-4\sqrt{37}}{14}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{37}}{2*7}=\frac{12+4\sqrt{37}}{14}
$