3x^2-14x-20=x^2-2(x+10)

Solution for 3x^2-14x-20=x^2-2(x+10) equation:

3x^2-14x-20=x^2-2(x+10)

We move all terms to the left:

3x^2-14x-20-(x^2-2(x+10))=0


We calculate terms in parentheses: -(x^2-2(x+10)), so:

x^2-2(x+10)

We multiply parentheses

x^2-2x-20

Back to the equation:

-(x^2-2x-20)


We get rid of parentheses

3x^2-x^2-14x+2x+20-20=0

We add all the numbers together, and all the variables

2x^2-12x=0

a = 2; b = -12; c = 0;
Δ = b2-4ac
Δ = -122-4·2·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-12}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+12}{2*2}=\frac{24}{4}
=6
$