2x^2-17x+42=x^2

Solution for 2x^2-17x+42=x^2 equation:

2x^2-17x+42=x^2

We move all terms to the left:

2x^2-17x+42-(x^2)=0

determiningTheFunctionDomain
2x^2-x^2-17x+42=0

We add all the numbers together, and all the variables

x^2-17x+42=0

a = 1; b = -17; c = +42;
Δ = b2-4ac
Δ = -172-4·1·42
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-17)-11}{2*1}=\frac{6}{2}
=3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-17)+11}{2*1}=\frac{28}{2}
=14
$