10p^2+22p-100=0

Solution for 10p^2+22p-100=0 equation:

10p^2+22p-100=0

a = 10; b = 22; c = -100;
Δ = b2-4ac
Δ = 222-4·10·(-100)
Δ = 4484
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:

$\sqrt{\Delta}=\sqrt{4484}=\sqrt{4*1121}=\sqrt{4}*\sqrt{1121}=2\sqrt{1121}$
$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{1121}}{2*10}=\frac{-22-2\sqrt{1121}}{20}
$
$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{1121}}{2*10}=\frac{-22+2\sqrt{1121}}{20}
$