12x^2-20x+10=3x+5

Solution for 12x^2-20x+10=3x+5 equation:

12x^2-20x+10=3x+5

We move all terms to the left:

12x^2-20x+10-(3x+5)=0

We get rid of parentheses

12x^2-20x-3x-5+10=0

We add all the numbers together, and all the variables

12x^2-23x+5=0

a = 12; b = -23; c = +5;
Δ = b2-4ac
Δ = -232-4·12·5
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{-(-23)-17}{2*12}=\frac{6}{24}
=1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+17}{2*12}=\frac{40}{24}
=1+2/3
$