2x^2-7x+16=x^2-6x+58

Solution for 2x^2-7x+16=x^2-6x+58 equation:

2x^2-7x+16=x^2-6x+58

We move all terms to the left:

2x^2-7x+16-(x^2-6x+58)=0

We get rid of parentheses

2x^2-x^2-7x+6x-58+16=0

We add all the numbers together, and all the variables

x^2-1x-42=0

a = 1; b = -1; c = -42;
Δ = b2-4ac
Δ = -12-4·1·(-42)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-13}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+13}{2*1}=\frac{14}{2}
=7
$