2p^2+18p=28

Solution for 2p^2+18p=28 equation:

2p^2+18p=28

We move all terms to the left:

2p^2+18p-(28)=0

a = 2; b = 18; c = -28;
Δ = b2-4ac
Δ = 182-4·2·(-28)
Δ = 548
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{548}=\sqrt{4*137}=\sqrt{4}*\sqrt{137}=2\sqrt{137}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{137}}{2*2}=\frac{-18-2\sqrt{137}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{137}}{2*2}=\frac{-18+2\sqrt{137}}{4}
$