2x^2+3x-5=x^2+11x-17

Solution for 2x^2+3x-5=x^2+11x-17 equation:

2x^2+3x-5=x^2+11x-17

We move all terms to the left:

2x^2+3x-5-(x^2+11x-17)=0

We get rid of parentheses

2x^2-x^2+3x-11x+17-5=0

We add all the numbers together, and all the variables

x^2-8x+12=0

a = 1; b = -8; c = +12;
Δ = b2-4ac
Δ = -82-4·1·12
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{-(-8)-4}{2*1}=\frac{4}{2}
=2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4}{2*1}=\frac{12}{2}
=6
$