2p^2+7p=14

Solution for 2p^2+7p=14 equation:

2p^2+7p=14

We move all terms to the left:

2p^2+7p-(14)=0

a = 2; b = 7; c = -14;
Δ = b2-4ac
Δ = 72-4·2·(-14)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-\sqrt{161}}{2*2}=\frac{-7-\sqrt{161}}{4}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+\sqrt{161}}{2*2}=\frac{-7+\sqrt{161}}{4}
$