3x^2+4x-10=2(x^2-x+3)

Solution for 3x^2+4x-10=2(x^2-x+3) equation:

3x^2+4x-10=2(x^2-x+3)

We move all terms to the left:

3x^2+4x-10-(2(x^2-x+3))=0


We calculate terms in parentheses: -(2(x^2-x+3)), so:

2(x^2-x+3)

We multiply parentheses

2x^2-2x+6

Back to the equation:

-(2x^2-2x+6)


We get rid of parentheses

3x^2-2x^2+4x+2x-6-10=0

We add all the numbers together, and all the variables

x^2+6x-16=0

a = 1; b = 6; c = -16;
Δ = b2-4ac
Δ = 62-4·1·(-16)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-10}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+10}{2*1}=\frac{4}{2}
=2
$