3x^2-9x+7=2x^2-4x+13

Solution for 3x^2-9x+7=2x^2-4x+13 equation:

3x^2-9x+7=2x^2-4x+13

We move all terms to the left:

3x^2-9x+7-(2x^2-4x+13)=0

We get rid of parentheses

3x^2-2x^2-9x+4x-13+7=0

We add all the numbers together, and all the variables

x^2-5x-6=0

a = 1; b = -5; c = -6;
Δ = b2-4ac
Δ = -52-4·1·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-7}{2*1}=\frac{-2}{2}
=-1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+7}{2*1}=\frac{12}{2}
=6
$