3x^2+23x+12=32

Solution for 3x^2+23x+12=32 equation:

3x^2+23x+12=32

We move all terms to the left:

3x^2+23x+12-(32)=0

We add all the numbers together, and all the variables

3x^2+23x-20=0

a = 3; b = 23; c = -20;
Δ = b2-4ac
Δ = 232-4·3·(-20)
Δ = 769
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-\sqrt{769}}{2*3}=\frac{-23-\sqrt{769}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+\sqrt{769}}{2*3}=\frac{-23+\sqrt{769}}{6}
$