2x^2=48+3x

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

2x^2=48+3x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

2x^2-3x-48=0

a = 2; b = -3; c = -48;
Δ = b2-4ac
Δ = -32-4·2·(-48)
Δ = 393
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{-(-3)-\sqrt{393}}{2*2}=\frac{3-\sqrt{393}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{393}}{2*2}=\frac{3+\sqrt{393}}{4}
$