7x^2+48=300

Solution for 7x^2+48=300 equation:

7x^2+48=300

We move all terms to the left:

7x^2+48-(300)=0

We add all the numbers together, and all the variables

7x^2-252=0

a = 7; b = 0; c = -252;
Δ = b2-4ac
Δ = 02-4·7·(-252)
Δ = 7056
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{7056}=84$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-84}{2*7}=\frac{-84}{14}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+84}{2*7}=\frac{84}{14}
=6
$