12x+3(10x-7)=-12x+20+5x2

Solution for 12x+3(10x-7)=-12x+20+5x2 equation:

12x+3(10x-7)=-12x+20+5x^2

We move all terms to the left:

12x+3(10x-7)-(-12x+20+5x^2)=0

We multiply parentheses

-(-12x+20+5x^2)+12x+30x-21=0

We get rid of parentheses

-5x^2+12x+12x+30x-20-21=0

We add all the numbers together, and all the variables

-5x^2+54x-41=0

a = -5; b = 54; c = -41;
Δ = b2-4ac
Δ = 542-4·(-5)·(-41)
Δ = 2096
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:

$\sqrt{\Delta}=\sqrt{2096}=\sqrt{16*131}=\sqrt{16}*\sqrt{131}=4\sqrt{131}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(54)-4\sqrt{131}}{2*-5}=\frac{-54-4\sqrt{131}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(54)+4\sqrt{131}}{2*-5}=\frac{-54+4\sqrt{131}}{-10}
$