3x^2+12=7x+103x2+12=7x+10

Solution for 3x^2+12=7x+103x2+12=7x+10 equation:

3x^2+12=7x+103x^2+12=7x+10

We move all terms to the left:

3x^2+12-(7x+103x^2+12)=0

We get rid of parentheses

3x^2-103x^2-7x-12+12=0

We add all the numbers together, and all the variables

-100x^2-7x=0

a = -100; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·(-100)·0
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*-100}=\frac{0}{-200}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*-100}=\frac{14}{-200}
=-7/100
$