3x^2+8x=300

Solution for 3x^2+8x=300 equation:

3x^2+8x=300

We move all terms to the left:

3x^2+8x-(300)=0

a = 3; b = 8; c = -300;
Δ = b2-4ac
Δ = 82-4·3·(-300)
Δ = 3664
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:

$\sqrt{\Delta}=\sqrt{3664}=\sqrt{16*229}=\sqrt{16}*\sqrt{229}=4\sqrt{229}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{229}}{2*3}=\frac{-8-4\sqrt{229}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{229}}{2*3}=\frac{-8+4\sqrt{229}}{6}
$