3x+48=x^2+7x-3

Solution for 3x+48=x^2+7x-3 equation:

3x+48=x^2+7x-3

We move all terms to the left:

3x+48-(x^2+7x-3)=0

We get rid of parentheses

-x^2+3x-7x+3+48=0

We add all the numbers together, and all the variables

-1x^2-4x+51=0

a = -1; b = -4; c = +51;
Δ = b2-4ac
Δ = -42-4·(-1)·51
Δ = 220
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{220}=\sqrt{4*55}=\sqrt{4}*\sqrt{55}=2\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{55}}{2*-1}=\frac{4-2\sqrt{55}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{55}}{2*-1}=\frac{4+2\sqrt{55}}{-2}
$