2x^2+28x=200

Solution for 2x^2+28x=200 equation:

2x^2+28x=200

We move all terms to the left:

2x^2+28x-(200)=0

a = 2; b = 28; c = -200;
Δ = b2-4ac
Δ = 282-4·2·(-200)
Δ = 2384
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:

$\sqrt{\Delta}=\sqrt{2384}=\sqrt{16*149}=\sqrt{16}*\sqrt{149}=4\sqrt{149}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-4\sqrt{149}}{2*2}=\frac{-28-4\sqrt{149}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+4\sqrt{149}}{2*2}=\frac{-28+4\sqrt{149}}{4}
$