3x^2-2x+10=x^2+2(x+5)

Solution for 3x^2-2x+10=x^2+2(x+5) equation:

3x^2-2x+10=x^2+2(x+5)

We move all terms to the left:

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


We calculate terms in parentheses: -(x^2+2(x+5)), so:

x^2+2(x+5)

We multiply parentheses

x^2+2x+10

Back to the equation:

-(x^2+2x+10)


We get rid of parentheses

3x^2-x^2-2x-2x-10+10=0

We add all the numbers together, and all the variables

2x^2-4x=0

a = 2; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·2·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*2}=\frac{0}{4}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*2}=\frac{8}{4}
=2
$