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

Solution for -2x^2-x+10=-3x^2+6x equation:

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

We move all terms to the left:

-2x^2-x+10-(-3x^2+6x)=0

We add all the numbers together, and all the variables

-2x^2-(-3x^2+6x)-1x+10=0

We get rid of parentheses

-2x^2+3x^2-6x-1x+10=0

We add all the numbers together, and all the variables

x^2-7x+10=0

a = 1; b = -7; c = +10;
Δ = b2-4ac
Δ = -72-4·1·10
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-3}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+3}{2*1}=\frac{10}{2}
=5
$