2^2-7x+4=x^2-8

Solution for 2^2-7x+4=x^2-8 equation:

2^2-7x+4=x^2-8

We move all terms to the left:

2^2-7x+4-(x^2-8)=0

We add all the numbers together, and all the variables

-7x-(x^2-8)+8=0

We get rid of parentheses

-x^2-7x+8+8=0

We add all the numbers together, and all the variables

-1x^2-7x+16=0

a = -1; b = -7; c = +16;
Δ = b2-4ac
Δ = -72-4·(-1)·16
Δ = 113
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-\sqrt{113}}{2*-1}=\frac{7-\sqrt{113}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{113}}{2*-1}=\frac{7+\sqrt{113}}{-2}
$