7p^2+6p-3=0

Solution for 7p^2+6p-3=0 equation:

7p^2+6p-3=0

a = 7; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·7·(-3)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{30}}{2*7}=\frac{-6-2\sqrt{30}}{14}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{30}}{2*7}=\frac{-6+2\sqrt{30}}{14}
$