3x^2/4=x+12

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

3x^2/4=x+12

We move all terms to the left:

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

We get rid of parentheses

3x^2/4-x-12=0

We multiply all the terms by the denominator


3x^2-x*4-12*4=0

We add all the numbers together, and all the variables

3x^2-x*4-48=0

Wy multiply elements

3x^2-4x-48=0

a = 3; b = -4; c = -48;
Δ = b2-4ac
Δ = -42-4·3·(-48)
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{-(-4)-4\sqrt{37}}{2*3}=\frac{4-4\sqrt{37}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{37}}{2*3}=\frac{4+4\sqrt{37}}{6}
$