8x^2+16x=300

Solution for 8x^2+16x=300 equation:

8x^2+16x=300

We move all terms to the left:

8x^2+16x-(300)=0

a = 8; b = 16; c = -300;
Δ = b2-4ac
Δ = 162-4·8·(-300)
Δ = 9856
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:

$\sqrt{\Delta}=\sqrt{9856}=\sqrt{64*154}=\sqrt{64}*\sqrt{154}=8\sqrt{154}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{154}}{2*8}=\frac{-16-8\sqrt{154}}{16}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{154}}{2*8}=\frac{-16+8\sqrt{154}}{16}
$