3x^2+4x+12=100

Solution for 3x^2+4x+12=100 equation:

3x^2+4x+12=100

We move all terms to the left:

3x^2+4x+12-(100)=0

We add all the numbers together, and all the variables

3x^2+4x-88=0

a = 3; b = 4; c = -88;
Δ = b2-4ac
Δ = 42-4·3·(-88)
Δ = 1072
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:

$\sqrt{\Delta}=\sqrt{1072}=\sqrt{16*67}=\sqrt{16}*\sqrt{67}=4\sqrt{67}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{67}}{2*3}=\frac{-4-4\sqrt{67}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{67}}{2*3}=\frac{-4+4\sqrt{67}}{6}
$