3x^2+20x+12=10

Solution for 3x^2+20x+12=10 equation:

3x^2+20x+12=10

We move all terms to the left:

3x^2+20x+12-(10)=0

We add all the numbers together, and all the variables

3x^2+20x+2=0

a = 3; b = 20; c = +2;
Δ = b2-4ac
Δ = 202-4·3·2
Δ = 376
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:

$\sqrt{\Delta}=\sqrt{376}=\sqrt{4*94}=\sqrt{4}*\sqrt{94}=2\sqrt{94}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{94}}{2*3}=\frac{-20-2\sqrt{94}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{94}}{2*3}=\frac{-20+2\sqrt{94}}{6}
$