2x+300=5x^2

Solution for 2x+300=5x^2 equation:

2x+300=5x^2

We move all terms to the left:

2x+300-(5x^2)=0

determiningTheFunctionDomain
-5x^2+2x+300=0

a = -5; b = 2; c = +300;
Δ = b2-4ac
Δ = 22-4·(-5)·300
Δ = 6004
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:

$\sqrt{\Delta}=\sqrt{6004}=\sqrt{4*1501}=\sqrt{4}*\sqrt{1501}=2\sqrt{1501}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{1501}}{2*-5}=\frac{-2-2\sqrt{1501}}{-10}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{1501}}{2*-5}=\frac{-2+2\sqrt{1501}}{-10}
$