3x^2+12=7x+32

Solution for 3x^2+12=7x+32 equation:

3x^2+12=7x+32

We move all terms to the left:

3x^2+12-(7x+32)=0

We get rid of parentheses

3x^2-7x-32+12=0

We add all the numbers together, and all the variables

3x^2-7x-20=0

a = 3; b = -7; c = -20;
Δ = b2-4ac
Δ = -72-4·3·(-20)
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-17}{2*3}=\frac{-10}{6}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+17}{2*3}=\frac{24}{6}
=4
$