3x^2-21x+48=-1+x^2

Solution for 3x^2-21x+48=-1+x^2 equation:

3x^2-21x+48=-1+x^2

We move all terms to the left:

3x^2-21x+48-(-1+x^2)=0

We get rid of parentheses

3x^2-x^2-21x+1+48=0

We add all the numbers together, and all the variables

2x^2-21x+49=0

a = 2; b = -21; c = +49;
Δ = b2-4ac
Δ = -212-4·2·49
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{-(-21)-7}{2*2}=\frac{14}{4}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+7}{2*2}=\frac{28}{4}
=7
$