(4x-2)(x+3)=-6x+20

Solution for (4x-2)(x+3)=-6x+20 equation:

(4x-2)(x+3)=-6x+20

We move all terms to the left:

(4x-2)(x+3)-(-6x+20)=0

We get rid of parentheses

(4x-2)(x+3)+6x-20=0

We multiply parentheses ..

(+4x^2+12x-2x-6)+6x-20=0

We get rid of parentheses

4x^2+12x-2x+6x-6-20=0

We add all the numbers together, and all the variables

4x^2+16x-26=0

a = 4; b = 16; c = -26;
Δ = b2-4ac
Δ = 162-4·4·(-26)
Δ = 672
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:

$\sqrt{\Delta}=\sqrt{672}=\sqrt{16*42}=\sqrt{16}*\sqrt{42}=4\sqrt{42}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-4\sqrt{42}}{2*4}=\frac{-16-4\sqrt{42}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+4\sqrt{42}}{2*4}=\frac{-16+4\sqrt{42}}{8}
$