1x^2=2x+48

Solution for 1x^2=2x+48 equation:

1x^2=2x+48

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

x^2-2x-48=0

a = 1; b = -2; c = -48;
Δ = b2-4ac
Δ = -22-4·1·(-48)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-14}{2*1}=\frac{-12}{2}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+14}{2*1}=\frac{16}{2}
=8
$